Critical numbers, intervals of increase and decrease, Relative extrema

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Here are some important points to note, while solving problems on this topic.

Critical points:

These are the points where first derivative of function is either 0 or undefined.

Interval of increase:

The interval where first derivative is positive. f'(x) >0 (derivative graph is above x axis)

Interval of decrease:

The interval where first derivative is negative. f'(x) <0 (derivative graph is below x axis)

Relative maxima:

The critical point where derivative f'(x) changes from + to –

Relative minima:

The critical point where derivative f'(x) changes from – to +

 

Example : Find critical points for the following function 

Step 1: Find first derivative using product rule.

Step 2: Set the derivative  equal to 0 and solve for x.

x=0 can’t be used as critical point because this is not in the domain of lnx as log functions are not defined for 0 and negative values.

,  rewriting the expression in terms of sqrt, we get

So the only critical point is

 

Reading the derivative graphs

This video explain how to read and interpret first derivative graph and get all information about intervals of increase, decrease and relative maxima, minima.

 

 

 

Example 2: Find the intervals where given function is increasing, decreasing and then find relative maxima and minima.
  on interval   

Step 1: Find first derivative of given function using chain rule.

Given function can be rewritten as,

Step 2: Set the first derivative equal to 0 and solve for x.

Using zero product rule, set each factor equal to 0 and solve.

-2sinx =0        or      cosx =0

sinx=0                       cosx =0

Sinx is 0 at 0 and each multiple of pi  and cosx is 0 at odd multiples of pi/2. So we write all values inside given domain [0, 2pi)

Using these critical points, we get number line divided into 4 subintervals. Then check each interval by picking any testing point from each of them and plugging them into first derivative.

For interval   . lets use pi/4 as testing point  and plugin it into g'(x)

For interval    lets use 3pi/4 as testing point and plugin it into g'(x)

For interval    lets use 5pi/4 as testing point and plugin it into g'(x)

For interval   lets use 7pi/4 as testing point and plugin it into g'(x)

Next write all the results on a number line.

We get the following information from this sign chart.

Intervals of increase:

Intervals of decrease:

Relative maxima:

Relative minima:

 

 

 

Example :The graph of 𝑔′(𝑥), the derivative of 𝑔, is continuous and differentiable for all real numbers. 𝑔′(𝑥) has horizontal tangents at 𝑥 = −3 and 𝑥 = 3.

a) On what intervals is g(x) decreasing? Give reasong for your answer.

b) At which value of x, does g(x) has relative maxima and minima? Explain your reasoning.

c) Write the intervals where the function is concave up and down.

 

 

Example : The graphs of g'(x) and h'(x) are given on the interval [-3,7]. Use these graph to find the following:

a)   Let . On what intervals is f(x) increasing?

b)   Let k(x) = h'(x). On what intervals is k(x) decreasing? Explain the reason.

 

 

 

 

Practice problems:

Find intervals of increase and decrease for  the following functions.

 

 

 

Answers:

  1. Increasing    Decreasing
  2. Increasing    Decreasing

Mean value theorem

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Statement: let f(x) be  a function defined on [a,b] such that:

i) it is continuous on [a,b]

ii) it is differentiable on (a,b)

Then , there exist a real number C (a,b) such that

               \dpi{120} \mathbf{f'(c)=\frac{f(b)-f(a)}{b-a}}

In other words mean value theorem is an existence theorem which guarantees that instant rate of change is equal to average rate of change some where on the given interval if the two conditions of continuity and differentiability are satisfied.

i.e IROC=AROC

Here are few examples to understand practical use of mean value theorem.

Example 1. A hot air balloon is launched into the air with a human pilot. The function h(t), which is a twice differentiable function, gives the height of balloon in ft at time t min. The table below gives different values of height h(t) at different times.

  • t
  • h(t)
  • 0
  • 0
  • 5
  • 45
  • 10
  • 35
  • 40
  • 125

a) Is there a time t on interval     when balloon is 50 ft in the air ? Explain your answer.

To answer this question, we have to use IVT (intermediate value theorem) which says if a function f(x) is continuous on closed interval [a,b] then it takes on every real value from f(a) to f(b). Here we have

That means this function h(t) take on every value from 45 to 35 on closed interval      but it doesn’t reach to a height of 50 ft. So, there is no such time when balloon reach a height of 50 ft on given time interval.

b) Is there a time on interval   when balloon attain a velocity of 3ft/min? Justify your answer.

Since given height function is continuous and differentiable on given interval, mean value theorem can be applied here. So we can find instant rate of change using average rate of change on given interval because MVT guarantees existence of a point when Instant rate is equal to average rate of change.

Hence proved that there exist a point ‘c’ on interval   such that 

There exist a time when balloon attain a velocity of 3 ft/min.

 

Example2. Let g(x) be a continuous function whose derivative is a piecewise linear function. The graph of derivative g'(x) on the interval   is shown below. Does the mean value theorem applied on this interval guarantee a value ‘c’ for   such that g”(c) is equal to average rate of change? Explain your answer.

One of the condition of mean value theorem is that, function should be differentiable on given interval. But here the derivative function graph has several corners (sharp edges) and we know that function is not differentiable at sharp edges at -1 and 2. So, Mean value theorem (MVT) can’t be applied here .

Therefore we can’t guarantee a value ‘c’ for   such that g”(c) is equal to average rate of change.

 

 

Example3 : Let f be a twice diffferentiable function such that f(2)=5 and f(5)=2. Let g be the function given by g(x)=f(f(x)).

a) Show that g'(2)=g'(5). Use this result to explain why there must be a value k for 2<k<5 such that g”(k)=0.

b) Let h(x) = f(x)-x. Explain why there must be a value r for 2<r<5 such that h(r)=0.

 

Example4: The functions f and g are differentiable for all real numbers and g is strictly increasing. The table below gives the value of functions and their derivatives at some values of x. The function h is given as h(x)=f(g(x))-6.

  • x
  • 1
  • 2
  • 3
  • 4
  • f(x)
  • 6
  • 9
  • 10
  • -1
  • f'(x)
  • 4
  • 2
  • -4
  • 3
  • g(x)
  • 2
  • 3
  • 4
  • 6
  • g'(x)
  • 5
  • 1
  • 2
  • 7

a) Explain why there must be a value r for 1<r<3 such that h(r) = -5

b)Explain why there must be a value c for 1<c<3 such that h'(c) = -5

 

 

Real life applications of mean value theorem:

Example5 : A marathoner ran 26 mile, city marathon in 2 hr and 15 min. Like all other runners, he started from a stationary position. During last 5 meters, his leg cramped. He fell down and had to roll across the finish line. Prove that atleast twice, the runner was running at exactly 11 mph.

 

 

Practice problems:

1.Using the Mean Value Theorem, find where the instantaneous rate of change for the given functions is equivalent to the average rate of change on the given intervals.

a.

b.

2. Find the equation of tangent line to the graph of f(x) = 2x+sinx+1 on the interval    at the point which is the solution to the mean value theorem. Confirm your answer using graphic calculator.

 

 

 

 

 

 

 

Answers:

1.       a) x= -3                           b)

 

2.

Improper integrals

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When either one or both limits of integration are infinity, then it is called improper integral. The interval of integration is said to be over an infinite interval. We will call these integrals convergent if the associated limit exists and is a finite number and divergent if the associated limit either doesn’t exist or is (plus or minus) infinity.

Check whether given integral function is convergent or not.

int_1^oo {(4x+1)}/{(x^3-2x^2+x)} dx

Given integral function:

int_1^oo {(4x+1)}/{(x^3-2x^2+x)} dx 

 

Derivative of integral function

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The Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration. It says :

If f(x) is continuous over an interval \dpi{120} \left [ a,b \right ] and the function F(x) is defined by

\dpi{120} F(x)=\int_{a}^{x}f(t) dt

Then \dpi{120} F'(x)= f(x) over closed interval \dpi{120} \left [ a,b \right ].

Lets work on some examples:

 

  1. Find derivative of the given function, using the fundamental theorem of Calculus.

\dpi{120} {\color{Red} g(x) =\int_{0}^{x}\sqrt{1+t^2} dt}

Taking derivative of both sides,

\dpi{120} g'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{x}\sqrt{1+t^2} dt

Applying fundamental theorem of calculus,  integral sign get eliminated with \dpi{120} \frac{\mathrm{d} }{\mathrm{d} x}  and we replace variable t with x and we get,

\dpi{120} g'(x) = \sqrt{1+x^2}

Derivative of integral function when upper limit is not just x, but it is some function of x.

Then we replace variable t with the upper bound function and take its integral alongwith. Lets look at an example.

 

 

2.    Use fundamental theorem of calculus to find derivatve of following integral function.

\dpi{120} {\color{Red} \int_{5}^{x^2 +9}sin(t^3)dt}

Taking derivative of given integral,

\dpi{120} \frac{\mathrm{d} }{\mathrm{d} x}\int_{5}^{x^2 +9} sin(t^3)dt=sin(x^2+9)^3 . (x^2+9)'

\dpi{120} = sin(x^2+9)^3 . 2x

\dpi{120} = 2x sin(x^2 +9)^3

 

Derivative of integral function when integrand is a function of two variables.

In such a case we start by taking derivative of whole fucntion using product rule and then apply FDC in next step for second part.

 

 

3. Use fundamental theorem of calculus to find derivatve of following integral function.

\dpi{120} {\color{Red} y=\int_{1}^{tanx}e^x \sqrt{1+t^3} dt}

First we separate the expressions in x and t.

\dpi{120} y= e^x \int_{1}^{tanx}\sqrt{1+t^3} dt

Taking derivative of both sides and using product rule.

\dpi{120} \frac{\mathrm{d} }{\mathrm{d} x}[y]= \frac{\mathrm{d} }{\mathrm{d} x}[e^x \int_{1}^{tanx}\sqrt{1+t^3} dt]

\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x}[e^x ] .\int_{1}^{tanx}\sqrt{1+t^3} dt + e^x\frac{\mathrm{d} }{\mathrm{d} x}\int_{1}^{tanx}\sqrt{1+t^3}dt               [ used product rule here]

\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= e^x \int_{1}^{tanx}\sqrt{1+t^3} dt + e^x\frac{\mathrm{d} }{\mathrm{d} x}\int_{1}^{tanx}\sqrt{1+t^3}dt

We observe that first part on right side is just the given original function so we can replace it with y. For second part on right side, we can apply fundamental theorem of Calculus. Replace t with tanx and take derivative of tanx alongwith.

\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= y + e^x\sqrt{1+(tanx)^3} . (tanx)'

\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= y + e^x\sqrt{1+(tanx)^3} . sec^2(x)

\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= y + e^x sec^2(x)\sqrt{1+(tanx)^3}

 

Derivative of an integral function when both the bounds are functions of x.

When lower bound is no more a constant, instead, both upper and lower bounds are functions of x, then we break  the integral into two parts in such a way that lower bound is a constant in both parts. We can break the integral at 0 and manipulate it further to get desired form. Here is an example :

 

4. Find derivative of the following integral function using fundamental theorem of Calculus.

{\color{Red} y=\int_{x}^{2x}sin(t^2)dt}

Here we can break the integral at any constant ‘c’. We can also use c as 0.

y=\int_{x}^{c}sin(t^2)dt +\int_{c}^{2x}sin(t^2)dt

Presently, it is not in right format. We need to interchange the bounds of first integral. Using property of definite integral we can interchange the bounds by putting a negative sign in front.

y=-\int_{c}^{x}sin(t^2)dt +\int_{c}^{2x}sin(t^2)dt

Taking derivative of both sides,

\frac{dy}{dx}=-\frac{\mathrm{d} }{\mathrm{d} x}\int_{c}^{x}sin(t^2)dt +\frac{\mathrm{d} }{\mathrm{d} x}\int_{c}^{2x}sin(t^2)dt

Applying fundamental theorem of calculus,  integral sign get eliminated with \dpi{120} \frac{\mathrm{d} }{\mathrm{d} x}  , variable t replaced with x in first integral and 2x in second integral. Also derivatives of x and 2x taken alongwith.

\frac{\mathrm{d} y}{\mathrm{d} x}=-sin(x)^2(x)'+ sin(2x)^2 (2x)'

\frac{\mathrm{d} y}{\mathrm{d} x}=-sin(x)^2*1+ sin(4x^2) (2)

\frac{\mathrm{d} y}{\mathrm{d} x}=-sin(x^2)+ 2sin(4x^2)

 

 

Practice problems:

Using second fundamental theorem of calculus, find derivative of the following integrals:

1. \int_{2}^{sinx}e^{t^2}dt

 

2. \int_{5}^{5x+3}\sqrt{1+t^2}dt

 

3.\int_{sinx}^{tanx}\sqrt{1-t^2} dt

 

 

 

Answers:

1. cosx. e^{sinx}

2. 5\sqrt{25x^2 +30x+10}

3. -cos^2x +sec^2x\sqrt{1-tan^2x}

 

Derivative of Inverse function

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If f is a function with inverse function  f^{-1} , then derivative of inverse function is given by the following formula :

\dpi{120} (f^{-1})'(x)= \frac{1}{f'(f^{-1}(x))}

This implies the derivative of \dpi{120} f^{-1} is the reciprocal of the derivative of f(x), with argument and value reversed.

 

This formula is an immediate consequence of the definition of an inverse function and the chain rule.

\dpi{120} f(f^{-1}(x))=x

Taking the derivative of both sides,

\dpi{120} \frac{\mathrm{d}[ f(f^{-1}(x))]}{\mathrm{d} x}= \frac{\mathrm{d} [x]}{\mathrm{d} x}

\dpi{120} f^{'}(f^{-1}(x))\times (f^{-1})'(x) =1

\dpi{120} (f^{-1})'(x) =\frac{1}{f^{'}(f^{-1}(x))}

Alternatively,

If y= g(x) is the inverse of f(x), then

\dpi{120} g'(x)= \frac{1}{f'(g(x))}

 

Lets look at an example.

 

 1. let \dpi{120} {\color{Red} f(x) = 5+x^2 +tan(\frac{\pi x}{2})} defined on interval \dpi{120} {\color{Red} -1\leq x\leq 1} . Then find \dpi{120} {\color{Red} f'(f^{-1}(5))}

First we need to find value of inverse function at 5. We observe that

\dpi{120} f(0)= 5+0^2+tan(\frac{\pi\times 0}{2} )

\dpi{120} f(0) = 5 +tan(0)

\dpi{120} f(0) = 5

\dpi{120} 0=f^{-1}(5)

Now we can  plugin this value into formula and solve further.

\dpi{120} (f^{-1})'(x) =\frac{1}{f^{'}(f^{-1}(x))}

\dpi{120} (f^{-1})'(5) =\frac{1}{f^{'}(f^{-1}(5))}

\dpi{120} (f^{-1})'(5) =\frac{1}{f^{'}(0)}

Next we find derivative of given function at x=0

\dpi{120} f(x )= 5+x^2 +tan(\frac{\pi x}{2})

\dpi{120} f'(x )= 0+ 2x +sec^2(\frac{\pi x}{2})(\frac{\pi }{2})           [  used  chain  rule  ]

\dpi{120} f'(x )= 2x +\frac{\pi }{2}sec^2(\frac{\pi x}{2})

\dpi{120} f'(0 )= 2(0) +\frac{\pi }{2}sec^2(\frac{\pi (0)}{2})

\dpi{120} f'(0)= 0+ \frac{\pi }{2}sec^2(0)

\dpi{120} f(0)= \frac{\pi }{2}

Finally we get the answer as,

\dpi{120} (f^{-1})'(5) =\frac{1}{f^{'}(0)}=\frac{1}{\frac{\pi }{2}}

\dpi{120} (f^{-1})'(5) ={\frac{2 }{\pi }}

 

Lets look at another example through this video lesson.

 

 

Now its time for some practice.

  1. Given \dpi{120} f(x) = 1+ln(x-2) . Find \dpi{120} (f^{-1})'(1).
  2. Given \dpi{120} f(x) = x^5 +2x^3 +7x+1. Find \dpi{120} (f^{-1})'(1)

Algebra of functions

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Algebra of functions means combining different functions using four major algebraic operations addition, subtraction, multiplication and division.

Suppose we want to find sum of two functions f and g (both as functions of same variable x), then we can write this operation as:

(f+g)(x) = f(x) + g(x)

Same way we can apply other operations too.

What is the domain of combined function?

Domain of combined function would be the intersection of domains of two individual functions. Lets assume domain of function f(x) is A and domain  of function g(x) is B then Domain of combined functions would be \dpi{100} A\cap B

But we have to be careful while writing the domain of quotient function f/g  as denominator of an expression can never have 0.

Lets summarize all the rules of combining functions  and writing their domain in a tabular form.

 

OperationFormulaDomain
Sum(f+g)(x) = f(x) + g(x)A\cap B
Difference (f-g)(x) = f(x) – g(x)A\cap B
Product (f*g)(x) = f(x) *g(x)A\cap B
Quotient (f/g)(x) = f(x)/g(x)\left \{ x\in A\cap B|g(x)\neq 0) \right \}

 

Lets work on some examples to understand the concept better,

 

1. Do the operations as directed for given functions f and g  and then find the domain. 

\dpi{100} {\color{Red} f(x) = x^2 +x , g(x)= x^2}

a. (f+g)(x) 

b. (f-g)(x) 

c. (f*g)(x)

d. (f/g)(x)

Solution:

Here f(x) and g(x) both functions are polynomials. We know that polynomials are defined for all real numbers.

Domain of f =  \dpi{100} \left ( -\infty ,\infty \right )

Domain of g = \dpi{100} \left ( -\infty ,\infty \right )

Intersection of both domains = \dpi{100} \left ( -\infty ,\infty \right )

a.  Using the sum formula (f+g)(x) = f(x) +g(x)

\dpi{100} (f+g)(x)= x^2 +x+x^2 = 2x^2 +x

Domain : \dpi{100} \left ( -\infty ,\infty \right )

b. Using difference formula (f-g)(x)= f(x) – g(x)

\dpi{100} (f-g)(x)= x^2 +x-x^2 = x

Domain : \dpi{100} \left ( -\infty ,\infty \right )

c. Using product formula (f*g)(x)= f(x)*g(x)

\dpi{100} (f\ast g)(x)= (x^2 +x)x^2 = x^2.x^2+x.x^2                   Distributed x^2

\dpi{100} (f*g)(x)= x^4 +x^3                                                     Combined like terms

Domain : \dpi{100} \left ( -\infty ,\infty \right )

d. Using quotient formula

\dpi{100} \frac{f}{g}(x) =\frac{f(x)}{g(x)}

\dpi{100} \frac{f}{g}(x)=\frac{x^2+x}{x^2}                                                            Plugin values of f(x) and g(x)

Using the fact that any function becomes undefined when its denominator is 0 , x^2 can’t be 0

That means x can’t be 0 here. Therefore,

Domain : \dpi{100} \left \{ x\in \mathbb{R}: x\neq 0 \right \} where  R is the set of real numbers.

 

2. Perform the following operations for given functions and find their domain.

\dpi{100} {\color{Red} f(x) =\sqrt{16-x^2} , g(x)=\sqrt{x^2-1}}

a. (f +g)(x)

b. (f*g)(x)

c. (f/g)(x)

Solution:

Lets first find domain of each function.

Since radical (square root) functions are not defined for negative values, so we set radicand (inner part) ≥0 and solve for x.

\dpi{100} f(x) = \sqrt{16-x^2}

16- x² ≥ 0

16 ≥ x²

Domain f : [-4,4]

\dpi{100} g(x) = \sqrt{x^2 -1}

x² -1 ≥ 0

x² ≥ 1

Domain g : \dpi{100} \left ( -\infty , -1 \right ]\cup [1, \infty )

Intersection of both domains is given as : \dpi{100} \left [ -4, -1 \right ]\cup \left [ 1,4 \right ]

a. \dpi{100} (f+g)(x) = f(x) +g(x)

\dpi{100} (f+g)(x)= \sqrt{16-x^2}+\sqrt{x^2-1}                This expression can’t be simplified more so we leave this answer like this.

Domain: \dpi{100} \left [ -4, -1 \right ]\cup \left [ 1,4 \right ]

 

b. \dpi{100} (f*g)(x) = f(x)*g(x)

\dpi{100} (f*g)(x)= \sqrt{16-x^2}\sqrt{x^2 -1}

\dpi{100} (f*g)(x) = \sqrt{(16-x^2)(x^2 -1)}                    If needed , inner expressions can be multiplied and simplified.

Domain: \dpi{100} \left [ -4, -1 \right ]\cup \left [ 1,4 \right ]

 

c.  \dpi{100} \frac{f}{g}(x)=\frac{f(x)}{g(x)}

\dpi{100} \frac{f}{g}(x)= \frac{\sqrt{16-x^2}}{\sqrt{x^2 -1}}                                            For rational function , denominator can’t be 0.

Since denomaintor √(x² -1) would be 0 at -1 and 1  that means -1 and 1 must be excluded from intersection of domains.

Domain: \dpi{100} \left [ -4, -1 \right )\cup \left ( 1,4 \right ]

 

X and Y Intercepts

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Lets first understand what are x and y intercepts.

X intercept :

These are the points, where any line or curve intersects or crosses the x-axis.

How to write x-intercepts?

X intercepts are written in the form of  ordered pairs as \dpi{100} \left ( x_{1},0 \right ),\left ( x_{2},0 \right ) or can also be written using separated commas as \dpi{100} x= x_{1},x_{2}.

Y Intercept :

These are the points, where any line or curve intersects or crosses the y-axis.

How to write y-intercept?

Y intercept too, can also be written two ways, either ordered pair form (0,y) or using separated commas.

Note : A linear equation ( a line graph) can’t have more than one x and y intercept because a line can intersect x or y axis atmost at one point.

                                         

How to find x and y intercept ?

Its easy!

For x intercept, we just replace y with 0 and solve for x.

For y intercept we replace x with 0 and solve for y.

So we just do the opposite. (Its easy for you to memorize it like this:)

 

Lets work on some examples to understand this process more.

 

1. Find x and y intercepts of linear equation 2x+3y= 12.

x-intercept :

Plugin y as 0 and solve for x.

2x+3(0) = 12

2x+0= 12

2x=12

x=6  In ordered pair form we can write it as (6,0)

y-intercept :

Plugin x as 0 and solve for y.

2(0) +3y = 12

0+3y =12

3y=12

y=4 In ordered pair form we can write it as (0,4)

 

2. Find x and y intercepts of quadratic equation   \dpi{100} {\color{Red} y=x^{2}-7x+12}

y-intercept:

Plugin  x as 0 and solve for y.

\dpi{100} y=0^{2}-7(0)+12

y= 12     (0,12)

x- intercept :

Plugin y as 0 and solve for x.

\dpi{100} 0=x^{2}-7x+12

Factoring the above equation, we get

0 = (x-3)(x-4)

x-3 = 0     x-4 = 0

x =3            x = 4

So this curve has two x intercepts  written in ordered pair as  (3,0) and (4,0)

 

 

Practice problems:

  1. Find x and y intercepts for linear equation x+3y = 15
  2. Find x and y intercepts of quadratic equation  \dpi{100} y=x^{2}-5x+6

Operations on Scientific Notations worksheet

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Addition and subtraction in scientific notation:

While performing basic arithmetic operations like addition, subtraction, multiplication or division on   numbers written in scientific notation, we need to be careful and should follow some rules to reach the correct answer.

While adding and subtracting the numbers in scientific notation, exponent part of all the numbers must be same. If it is not same , then first make it same and then combine the numbers.

Don’t forget to write the final answer in scientific notation.

When exponent part  is same-

Example1: Combine \dpi{120} {\color{Red} 4.52\times 10^{3}}   and  \dpi{120} {\color{Red} 8.3\times 10^{3}}

Solution: Here we see that exponent part of both numbers is same (\dpi{120} 10^{3}). So we just need to combine the coefficients of these two numbers.

\dpi{120} 4.52\times 10^{3} + \dpi{120} 8.3\times 10^{3}   =  \dpi{120} (4.52+8.3)\times 10^{3}

= \dpi{120} 12.82\times 10^{3}

As the coefficient part can’t exceed 10, so we need to convert this  result  into scientific notation.

=  \dpi{120} 1.282\times 10^{4}

 

When exponent part is not same-

Example2: Simplify the following expression:   \dpi{120} {\color{Red} 8\times 10^{8}- 5\times 10 ^{6}}

Solution: As the exponent part is not same, we make them same first.

Again you can do it two ways. Either lower the exponent  \dpi{120} 10^{8}  to \dpi{120} 10^{6}   or  you can raise the exponent \dpi{120} 10^{6}   to  \dpi{120} 10^{8}

First way:  raise the exponent \dpi{120} 10^{6}   to  \dpi{120} 10^{8}    

As we know, when decimal is moved to left, exponent is positive (increased). We need to raise the exponent by 2 so decimal is moved towards left by 2 places and we get \dpi{120} 5\times 10^{6} = 0.05\times 10^{2}\times 10 ^{6}

= \dpi{120} 0.05\times 10^{8}

\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}   =  \dpi{120} 8\times 10^{8}- 0.05\times 10 ^{8}

= \dpi{120} (8-0.05)\times 10^{8}

=  \dpi{120} 7.95\times 10^{8}

Second way:  Either lower the exponent  \dpi{120} 10^{8}  to \dpi{120} 10^{6}

When decimal is moved to right, exponent is negative (decreased). We need to lower the exponent by 2 so decimal is moved towards right by 2 places  and we get    \dpi{120} 8\times 10 ^{8} = 8\times 10^{2}\times 10^{6} = 800\times 10^{6}

\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}  = \dpi{120} 800\times 10^{6}- 5\times 10 ^{6}

=  \dpi{120} 795\times 10^{6}

But this result is not in scientific notation. To get it in scientific notation decimal should be placed after first significant digit.

So final answer is   \dpi{120} 7.95\times 10^{8}

 

Example3. Simplify the expression  \dpi{120} {\color{Red} 9\times 10^{5} + 6\times 10 ^{7}}

Solution:   \dpi{120} 9\times 10^{5} + 6\times 10 ^{7}  =   \dpi{120} 9\times 10^{5} + 6\times 10 ^{2}\times 10^{5}

=   \dpi{120} 9\times 10^{5} + 600\times 10^{5}

= \dpi{120} (9+600)\times 10^{5}

=  \dpi{120} 609 \times 10^{5}

Then we convert it into scientific notation and get answer as  \dpi{120} 6.09 \times 10^{7}

 

Example4. Simplify the expression  \dpi{120} {\color{Red} 9\times 10^{5} - 6\times 10 ^{7}}

Solution:   \dpi{120} 9\times 10^{5} - 6\times 10 ^{7}   =   \dpi{120} 9\times 10^{5} - 6\times 10 ^{2}\times 10^{5}

=   \dpi{120} 9\times 10^{5} - 600\times 10^{5}

=  \dpi{120} (9-600)\times 10^{5}

=    \dpi{120} -591 \times 10^{5}

Rewriting it as scientific notation we get the final answer as   \dpi{120} -5.91\times 10^{7}

 

 

Practice problems:

Simplify each problem and write the answer in scientific notation.

  1. \dpi{120} 1\times 10^{6} - 8\times 10 ^{4}
  2. \dpi{120} 7\times 10^{7} + 3\times 10 ^{4}
  3. \dpi{120} 4\times 10^{8} - 9\times 10 ^{7}
  4.  \dpi{120} 3\times 10^{-2} - 6\times 10 ^{-5}

 

Answers:

1) \dpi{120} 9.2\times 10^{5}

2) \dpi{120} 7.003\times 10 ^{7}

3) \dpi{120} 3.1\times 10^{8}

4) \dpi{120} 2.994\times 10^{-2}

Scientific and Standard Notations

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Scientific and standard Notations

When we think about writing very large or very small numbers in an easy and quicker way, scientific notation comes to our rescue.

Therefore, scientific notation is a way to represent very small or very large numbers. It is also easy to compare many such numbers when written in scientific notations

How scientific notation is written:

Scientific notation has two parts, first one is the coefficient and second part consists of base 10 and its exponent.

Coefficient is always a number between 1 and 10 . It may contain one or more than one significant digits with a decimal.

Base is always 10

Exponents is always an integer.

Converting from standard notation to scientific notation

When we move towards left, exponent is positive.

When we move towards right, exponent is negative.

Example1. Convert 450000 to scientific notation.

Solution: Since the coefficient must lie between 1 and 10 so decimal should be placed after 4. As decimal is moved 5 steps towards left we use positive exponent (5)

So the scientific notation is written as

\dpi{120} 4.5\times 10^{5}

 

Example2. Convert 0.000000215 to scientific notation.

Solution: This is very small number less than 1 so exponent would be negative.

Since coefficient must lie between 1 and 10 so we put decimal after first significant digit which is 2. For that we need to move decimal to 7 places right and whenever we move towards right while converting standard to scientific notation, we use negative exponents.

So the scientific notation is written as,

\dpi{120} 2.15\times 10^{-7}

 

Example3. Express 0.8 in scientific notation.

Solution: Since decimal should be placed after first significant digit so decimal is moved to right by only one place.

= \dpi{120} 8.0\times 10^{-1}

Or it can also be written as   \dpi{120} 8\times 10^{-1}

 

Example4. Expression 30 in scientific notation.

Solution: Coefficient is the first significant digit which is 3 here, so we need to move decimal towards left by one place.

= \dpi{120} 3.0\times 10^{1}

So 30 can be expressed in scientific notation as \dpi{120} 3\times 10^{1}

 

Converting from scientific notation to standard form

For given negative exponents we move towards left.

For given positive exponents we move towards right.

 

Example5. Convert \dpi{120} {\color{Red} 4\times 10^{-5}} to standard notation.

Solution: Since decimal is not given here so we assume 4 as 4.0 and because of negative exponent(-5) we move 5 places towards left.

\dpi{120} 4\times 10^{-5}  =

So the standard form of given number is 0.00004

 

Example6. Convert \dpi{120} {\color{Red} 3.27\times 10^{4}} to standard form.

Solution: Since exponents is given positive so we move decimal towards right by 4 places.

\dpi{120} 3.27\times 10^{4} =

So the standard form of given number is 32700.

 

Example7. Express \dpi{120} {\color{Red} 5.2\times 10^{0}} to standard form.

Solution: Since exponent is 0 that means we don’t need to move decimal. Moreover  \dpi{120} 10^{0} =1    so the given number remain same without any change in it.

\dpi{120} 5.2\times 10^{0}= 5.2\times 1= 5.2

 

 

Practice problems:

Express the given numbers in scientific notation.

  1. 850,000
  2. 0.0000917
  3. 5102000

Express the given numbers in standard form.

  1. \dpi{120} 4.28\times 10^{-2}
  2. \dpi{120} 3.162\times 10 ^{4}
  3. \dpi{120} 1.2\times 10 ^{-5}

 

Answers:

  1. \dpi{120} 8.5\times 10^{5}
  2. \dpi{120} 9.17\times 10^{-5}
  3. \dpi{120} 5.102\times 10 ^{6}
  4. 0.0428
  5. 31620
  6. 0.000012

 

Worksheet 2 (Differential Calculus)

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1.  If f(x)=|x-3| then f'(1) is

 a) 0

b) 1

c) -1

d) Non existent

Solution: Derivative of this absolute function can be found either by using definition of absolute value function or by using formula.

Lets first use definition of absolute value function.

\dpi{120} f(x) =\begin{cases} & x-3 \text{ if } x\geq 3 \\ & -(x-3) \text{ if } x < 3 \end{cases}

Since x=1 lie in second domain (x<3) so we use lower function.

f(x) = -(x-3)

f ‘(x) = -(1-0) =-1

Therefore f ‘(1) = -1. Option c is correct.

Other way: Using formula for derivative of absolute functions.

Let  f(x) = |U|  where U= g(x)

Then \dpi{120} f'(x)=\frac{U}{|U|}*U'     or    \dpi{120} f'(x)=\frac{g(x)}{|g'(x)|}*g'(x)

\dpi{120} f'(x)=\frac{(x-3)}{|x-3|}*(x-3)'

f'(1)=\frac{(1-3)}{|1-3|}*1 =\frac{-2}{|-2|} =\frac{-2}{2}=-1

 

2. Given that \dpi{120} {\color{Red} f(x) =x^{10}h(x)}   and h(-1)=4 , h'(-1) = 7 . Then find f ‘(-1).

a) 4

b) -33

c)-40

d)-70

Solution:  Since f(x) is a product of two functions x^10 and h(x) so we use product rule to find derivative f'(x)

f'(x)=(x^{10})'h(x)+x^{10}h'(x)

f'(x)=(10x^{9})'h(x)+x^{10}h'(x)

f'(-1)=10(-1)^{9}h(-1)+(-1)^{10}h'(-1)

Plugin given values of h(1) and h'(-1)

f'(-1)= -10(4)+ (1)(7)

f ‘(-1) = -33

Option b)  -33 is correct answer.

 

3.   If {\color{Red} f'(x) =\frac{(x+2)^2)(x^2 -1)}{3}}    and  {\color{Red} g(x)= f(\sqrt{x+1})}  what is the value of g'(3) ?

a) 16

b) 1/4

c) 4

d) 16

Solution : First we find derivative of g(x) using chain rule.

g'(x)= f'(\sqrt{x+1})* (\sqrt{x+1})'

g'(x)= f'(\sqrt{x+1})* \frac{1}{2\sqrt{x+1}}

g'(3)= f'(\sqrt{3+1})* \frac{1}{2\sqrt{3+1}}

g'(3)= f'(2)* \frac{1}{2*2}

To get f'(2) we plugin x=2 into given expression for f'(x)

f'(2) =\frac{(2+2)^2(2^2 -1)}{3}= 16

Using f'(2)= 16 we get g'(3) as

g'(3) = 16 *(1/4) = 4

g'(3)= 4

So option c is the answer.

4.  Functions f, g and h are twice differentiable functions with g(2)= h(2)= 4. The line  {\color{Red} y= 4+\frac{2}{3}(x-2)}   is tangent to both the graphs of g at x=2 and graph of h  at x=2.

a) find h'(2)

b) Let a be a function given as

{\color{Red} a(x)= 3x^3 h(x)}.  Write an expression for a'(x) and a'(2).

c) The function h satisfies that  {\color{Red} h(x)= \frac{x^2-4}{1-(f(x))^3} ; x \neq 2} . It is known that {\color{Red} \lim_{x\rightarrow 2}h(x)} can be evaluated using L Hospital’s  rule. Use  this limit to find f(2)  and f ‘(2).

d) It is known that g(x) ≤ h(x)  for 1<x<3. Let k be a function satisfying g(x)≤k(x)≤h(x) for 1<x<3. Is k continuous  at x=2? Justify your answer.

Solution: It is known that slope of tangent line to the curve at given point is same  the slope of curve at that point.

Slope of tangent line at x=2 is  2/3

and that would be the slope of curve h(x) at x=2. so h'(2) = 2/3.

b)  a(x)= 3x^3 h(x).  Since given function is product of two functions so using product rule,

a'(x)= 3[(x^3)' h(x) + x^3 h'(x)]

a'(x)= 3[(3x^2)' h(x) + x^3 h'(x)]

a'(2)= 3[(3(2)^2)' h(2) + 2^3 h'(2)]

a'(2)= 3[12*4 + 8*2/3]

a'(2) =  160

c)

\lim_{x\rightarrow 2}h(x)= \lim_{x\rightarrow 2} \frac{x^2-4}{1-(f(x))^3}

It is given that this limit can be  evaluated using L Hospital’s rule. Which means it would have intermediate form(0/0)  when x approaches to 2. so we have

\frac{2^2-4}{1-(f(2))^3} = \frac{0}{0}

1-(f(2))^3=0

(f(2))^3= 1 \Rightarrow f(2)= 1

Since h(x) is continuous  so we have,

\lim_{x\rightarrow 2}h(x)=h(2)

Using L Hospital’s rule to find its limit,

\lim_{x\rightarrow 2}h(x)= \lim_{x\rightarrow 2} \frac{(x^2-4)'}{(1-(f(x))^3)'} = \frac{2x-0}{0-3(f(x))^2 f'(x)} = 4

\frac{2x}{-3(f(x))^2 f'(x)} =4

\frac{2*2}{-3(f(2))^2 f'(2)} =4

\frac{4}{-3(1)^2 f'(2)} =4

\frac{1}{-3f'(2)} =1

-\frac{1}{3} = f'(2)

d) Given that

g(x)\leq k(x)\leq h(x) on 1<x<3

Since g(x) and h(x) are given continuous  and g(2)= h(2) = 4

\lim_{x\rightarrow 2}g(x) =4    and   \lim_{x\rightarrow 2}h(x) =4 . Using sandwich theorem  we can conclude that,

\lim_{x\rightarrow 2}k(x) =4  and k(2)= 4

Therefore K(x) is continuous at x=2.

 

5. Let {\color{Red} g(x)= \int_{2}^{x}f(t) dt}  for 0≤t≤7 where the graph of differentiable function f, is shown below.

a) Find g(3) , g'(3)  and g”(3)

Using the given definition of g(x) we have,

g(3)= \int_{2}^{3}f(t) dt    which means  the area under the curve f(t) from x=2 to x=3. This area can be found using geometric formulas.

Area under the curve for given x values  is a trapezoid, so we can use trapezoid’s area formula.

A = 1/2 (sum of parallel sides)* distance between parallel sides.

g(3) = A= 1/2(2+4)*1 = 6/2 = 3

Using Fundamental Theorem of  Calculus

g'(3) = f(3) = 2

And g”(3)= f ‘(3)  which is slope of tangent line at x=3

slope = (4-0)/(2-4) = -2

so  g”(3) = -2

b) Find average rate of change of g(x) on 0≤c≤3

Average rate = \frac{g(3)-g(0)}{3-0}

For that first we need to find g(0) , same way as we had found g(3).

g(0)= \int_{2}^{0}f(t) dt

Which can be rewritten using property of definite integral.

g(0)= -\int_{0}^{2}f(t) dt . which  represent the area under the curve from 0 to 2.

g(0) =  – [(1/2)*2*4] = -4

Average rate = \frac{g(3)-g(0)}{3-0} = \frac{3-(-4)}{3} = \frac{7}{3}

c) For how many values of c, where 0<c<3, is g'(c) equal to average rate found in part b? Justify your answer.

We know that ,

g(c)= \int_{2}^{c}f(t) dt

then g'(c) = f(c) = 7/3

And the function f(t) gets the value 7/3 twice on the interval 0<c<3.

d) Find the Graph of gx coordinate of each point of inflection of the graph of g on the interval 0<t<7. Justify your answer.

We know that g'(x) = f(x) and g”(x)= f'(x).

Graph of g will have inflection points where f(x) changes its direction. We can observe from given graph of f that it changes direction from increasing to decreasing at x=2  and again from decreasing to increasing at x=5. Also slopes f ”(2) and f”(5) are undefined being sharp corners. So the two inflections points are at x=2,5.

 

6.  For what values of a and c , is the piecewise function differentiable.

{\color{Red} f(x)=\begin{cases} & ax^2+sinx, \text{ if } x\leq \pi \\ & 2x-c , \text{ if } x > \pi \end{cases} }

We know that functions ax^2+sinx  and 2x-c  are continuous and differentiable in their respective domains.

f'(x)=\begin{cases} & 2ax+cosx, \text{ if } x\leq \pi \\ & 2 , \text{ if } x > \pi \end{cases}

We are given that piecewise function is differentiable at x=π too. That means left and right hand derivatives of given function would be same at x=π. Therefore,

f'(\pi )=\begin{cases} & 2a \pi +cos(\pi ) , \text{ if } x\leq \pi \\ & 2 , \text{ if } x > \pi \end{cases}

2aπ +cos(π) = 2

2aπ -1 = 2

2a π = 3

a = \frac{3}{2\pi }

Also piecewise function is continuous at pi so left and right hand limits would be same as f(π)

Left limit is,

\lim_{x\rightarrow \pi^- }(ax^2+sinx)= a\pi ^2+sin(\pi )= a\pi ^2   which is same as value of function , i.e f(π)

And right limit is ,

\lim_{x\rightarrow \pi^+ }2x-c =2\pi -c

For the  function to be continuous. we set left and right limits  equal ,

a\pi ^2 =2\pi -c

c= 2\pi -a\pi ^2

Plugin value of a = 3/2π , we get

c=2\pi -\frac{3}{2\pi }*\pi ^2

c=2\pi -\frac{3\pi }{2}

c=\frac{\pi }{2}

So the solution is

a=\frac{3}{2\pi } , c=\frac{\pi }{2}

 

7. When the height of a cylinder is 12 cm and the radius is 4cm, the  circumference of the cylinder is increasing at a rate of  {\color{Red} \frac{\pi }{4}} cm/min, and the height of the cylinder is increasing four times faster than the radius. How fast is the volume of the cylinder changing?

Given:  height of cylinder(h)= 12 cm

radius of cylinder(r) = 4 cm

Rate of increase of circumference (2πr)

2\pi \frac{dr}{dt}=\frac{\pi }{4}

2 \frac{dr}{dt}=\frac{1 }{4} \Rightarrow \frac{dr}{dt}=\frac{1 }{8}

Height of cylinder is increasing 4 times faster than rate of change of radius.

\frac{dh}{dt} = 4*\frac{1}{8} =\frac{1}{2}

To find : Rate of change of volume  \dpi{120} \frac{dV}{dt} =?

Volume of cylinder is given as,

\dpi{120} V= \pi r^2 h

Differentiating both sides using product rule, with respect to time(t) ,

\dpi{120} \frac{dV}{dt}=\pi \left ( 2r\frac{dr}{dt}*h+r^2\frac{dh}{dt} \right )

Plugin value into this  expression.

\dpi{120} \frac{dV}{dt}=\pi \left ( 2(4)\frac{1}{8}*12+4^2*\frac{1}{2} \right )

\dpi{120} \frac{dV}{dt}=\pi \left ( 1*12+8 \right ) = 20\pi

So the volume of cylinder  is changing at the rate of 20π  cubic cm/min.

 

8. Let f be a function that is differentiable on open interval (1,10),  If, f(1) = -6 ,f(5)= 2 ,  and f(10) = -4. Which of the following must be true. Justify your answer.

i. f has atleast two zeros.

ii. The graph of f has atleast one horizontal tangent.

iii. f ‘(5) =0

iv. For some c, 1<c<10  , f(c) =1

v. For some c, 1<c<10 , f ‘(c)= 2/9

i, ii, iv and v must be true. Here is the reason for each.

i. f has atleast two zeros.

When  we move from left to right on interval (1,10) , y values goes from negative to positive (-6 to 2 )  and then again from positive to negative (from 2 to -4)  that means  function’s graph will cross x axis atleast twice. Which proves f has atleast two zeros.

ii. Since function’s graph cross x axis twice so graph will turn atleast once  and  there is  horizontal tangent at each turn. This proves  there is atleast one turn  and hence atleast one horizontal  tangent.

iv. Since the given function is differentiable and hence continuous. This graph is going from -6 to 2  when x is changing from 1 to 5 , that means continuous graph will pass through y value 1, while going from -6 to 2. So there must be some x value c  on given interval  such that f(c)=1

v. Since function is given differentiable and hence continuous on interval (1,10) , it satisfy hypothesis of Mean Value Theorem. That means there exist a number c  on interval (1,10) which satisfy conclusion of MVT.

Therefoere,

\dpi{120} f'(c)=\frac{f(10)-f(1)}{10-1}

\dpi{120} f'(c)=\frac{-4+6}{9}= \frac{2}{9}

which proves the statement  “ For some c, 1<c<10 , f ‘(c)= 2/9″

 

9.  Graph of velocity function of a particle is shown below, where t is measured in seconds. Find the intervals when particle is speeding up  and slowing down.

First we observe the given graph. We see that particle start at t=0 , reach max velocity at t=1 .

We make the following observations  by looking at the graph :-

Acceleration is positive as long as velocity graph is increasing and acceleration is negative as long as velocity graph is decreasing because acceleration is nothing but the slope of velocity.

Velocity is positive as long as graph is above x axis and velocity is negative as long as graph is below x axis.

Velocity is 0 at t=0 and t=2 seconds. It reaches to max value at t=1. So we make the following chart.

 

  • 0<t<1
  • Velocity (V)
  • Acceleration (A)
  • 0<t<1
  • +
  • +
  • 1<t<2
  • +
  • –
  • 2<t<3
  • –
  • –

Rules to follow :

Particle is speeding up when  velocity and acceleration both have same signs. Particle is slowing down when they have opposite signs.

So the answer is :

Particle is speeding up on intervals : (0,1)  and (2,3)

slowing down on interval : (1,2)

 

10 . A particle moves according to position function {\color{Red} S(t)= \frac{t}{1+t^2}} on the interval   t ≥0 where t is measured in seconds and S in feet.

a) Find when particle is at rest.

Particle is at rest when its velocity  is 0. So we find velocity function by taking its derivative  and then set velocity function =0 and solve for t.

V(t)=S'(t)= \frac{(1+t^2)t' - t(1+t^2)'}{(1+t^2)^2}

V(t)= \frac{(1+t^2)1 - t(2t)}{(1+t^2)^2} =\frac{1+t^2 -2t^2}{(1+t^2)^2}

V(t) =\frac{1-t^2}{(1+t^2)^2}

V(t) =0 \Rightarrow \frac{1-t^2}{(1+t^2)^2}=0

This expression would be 0 when its numerator is 0 so,

1-t^2 =0

t = -1,1

But t=-1  doesn’t lie in the given domain so we accept only t=1.

That means particle is at rest at t=1 second

 

b) Find the intervals for which particle is moving in positive(right) and negative(left) directions.

Particle is moving towards right when velocity is positive  and moving towards left when velocity is negative. So we need to check the intervals where velocity is positive  and where it is negative. Domain is given [0,∞)  which is divided into two intervals by t=1 (when velocity is 0) So we get the two intervals as,

[0,1) and (1,∞)

For interval [0,1)  We get  V(0.5) > 0

Velocity is positive, that means particle is moving towards right on [0,1)

For interval (1,∞) ,  We get V(2) < 0

Velocity is negative, that means particle is moving towards left on (1,∞)

 

c) Find the total distance traveled in first 8 seconds.

To find total distance, we find distance traveled  in positive direction as well as in negative direction.

Distance traveled in positive direction:

\left | S(1)-S(0) \right |= \frac{1}{2} -0 = \frac{1}{2} ft.

Distance traveled in negative direction:

\left | S(8)-S(1) \right |= \frac{1}{2} -\frac{8}{65}

Total Distance :

\frac{1}{2}+\frac{1}{2}-\frac{8}{65} = 1-\frac{8}{65} =\frac{57}{65} ft

 

d) When the particle is speeding up and slowing down? 

To find  when the particle is speeding up or slowing down, we need to find acceleration function. We know that acceleration is derivative of velocity function. so,

a(t)= V'(t)=\frac{(1+t^2)^2(1-t^2)'-((1+t^2)^2)'(1-t^2)}{(1+t^2)^4}

a(t)= \frac{(1+t^2)^2(-2t)-2(1+t^2)(2t)(1-t^2)}{(1+t^2)^4}

a(t)= \frac{(1+t^2)[(-2t)(1+t^2)-2(2t)(1-t^2)]}{(1+t^2)^4}

a(t)= \frac{[(-2t)(1+t^2)-2(2t)(1-t^2)]}{(1+t^2)^3}=\frac{-2t-2t^3-4t+4t^3}{(1+t^2)^3}

a(t)= \frac{-6t+2t^3}{(1+t^2)^3}=\frac{2t(t^2-3)}{(1+t)^3}

Setting a(t)=0  and solving for t, we get

2t(t^2-3)=0

t=0, t=\pm \sqrt{3}

But -√3 doesn’t lie in the given domain, so we have only t=0 and t=√3 .

Using t=√3 , we get the domain t≥0 divided into two intervals as [0,√3)  and (√3,∞). Now we check each interval one by one.

For interval [0,√3) we get ,

a(1)=\frac{2*1(1^2-3)}{(1+1)^3} = \frac{-1}{2} < 0

Acceleration is negative during this interval.

For interval (√3,∞)

a(2)=\frac{2*2(2^2-3)}{(1+2)^3} = \frac{4}{27} > 0

Acceleration is positive during this interval.

Lets make a sign chart using velocity and acceleration results,

  • 0<t<1
  • Velocity (v)
  • Acceleration(a)
  • 0<t<1
  • +
  • –
  • 1<t<√3
  • –
  • –
  • t>√3
  • –
  • +

Using the fact that particle will be speeding up when both velocity and acceleration have same signs (either both positive or both negative) and speeding down when both have different signs, we get the following results.

Speeding up on 1<t<√3

Speeding down on 0<t<1  and t>√3.

 

 

Practice Worksheet

  1. Find derivative of given function:f(x)= 2sin^{-1}(4x^3)
  2. Find second derivative of  y^2-x^2=9
  3. Find x values of the horizontal tangents of the curve   y= x^4 -18x^2-4
  4. Given the function f(x)= ax-bx^2 . Determine the values of a and b such that the line 3y-6x+6=0 is tangent to the function at x=3.
  5. At time t, the position of a body moving along the x-axis is given as  x(t)= t^3 -21t^2 +144t . Find the body’s acceleration
    each time the velocity is zero.
  6. The graphs of f and g are shown below. Use them to evaluate

a) \left ( fg \right )'(5)       b) \frac{d}{dx}\left ( \frac{g(x)}{f(x)} \right )_{x=5}

7.  A particle moves along a coordinate axis in the positive direction to the right. Its position at time t is given by s(t)=t^3-4t+2. Find v(1) and a(1) and use these values to answer the following questions.

a) Is the particle moving from left to right or from right to left at time t=1?

b) Is the particle speeding up or slowing down at time

8. The velocity-time graph of an object moving along a straight line is shown below. Find the intervals for which object is slowing down.

9. Find least  value of a such that the function f(x)= x^2+ax+1 is increasing on [1,2].

10. A man is walking at the rate of 6.5 km/hr towards the foot of a tower 120 m high. At what rate  is he approaching  the top of tower when he is 50m away from tower?

 

 

 

 

 

 

 

Answers:

  1. \frac{24x^2}{\sqrt{1-16x^6}}
  2. \frac{9}{y^3}
  3. x=-3, 0, 3
  4.  a= 2/3 , b=-2/9
  5. a(6)= -6m/s^2  and a(8)=6m/s^2
  6. a) 1/6    b)7/24
  7.  a) Right to left     b) slowing down
  8.  (1,3)
  9.  -2
  10. 2.5 km/hr