Subtopic Archive - Celestial Tutors

How limits help us to handle change at an instant

July 22, 2025/by admin_ct

Relationship between AROC and IROC

Average rate of change (AROC), which is represented by secant line passing through two points on a curve gives an approximation of rate of change on an interval. The formula to calculate AROC is very similar to slope formula which you have studied earlier to find slope of straight lines.

$AROC=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

But if we need to find rate of change at an instant then this interval $\left[x_1,x_2\right]$ is made smaller and smaller in such a way that difference between two end points of the interval is negligible. In this situation, if we draw a tangent line on this extremely small interval which is almost a point , then slope of this tangent line would represent instant rate of change(IROC).

We can represent this instant rate of change with the help of limits as:

$IROC=\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

Where h represent the distance between end points of extremely small interval.

Here is video tutorial on this topic.

Absolute extrema, Concavity and Points of Inflection

February 12, 2025/by admin_ct

Extreme Value Theorem:

If f(x) is continuous on closed interval [a,b] then function f has both an absolute minimum and absolute maximum value in the interval.

While the extreme value theorem guarantees the existence of absolute maximum and minimum value, it doesn’t tell us where they are. Follwoing steps are used to find absolute extrema for a function:

Find first derivative and set it equal to 0 to find critical values.
Plugin critical points and end points of given interval into original function and find the values at these points.
Greatest of all these values would be absolute maximum and smallest of these values would be the absolute minimum.

Note : Absolute maximum or minimum are also called global maximum or minimum.

Example 1: Find maximum and minimum values of k(x) on interval $\left[0,2\pi\right]$

$k(x)=e^{sinx}$

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

$k'(x)=e^{sinx}*(sinx)'$

$k'(x)=e^{sinx}(cosx)$

Step 2: Set the derivative =0 and solve for x to get critical points.

$e^{sinx}cosx=0$

Using zero product rule, either exponential part would be 0 or cosx would be 0. But exponential function is never 0 so we set only cosx equal to 0.

$cosx=0$

$x=\frac{\pi}{2},\frac{3\pi}2{}$

Step 3: Plugin these two critical points as well as end points of given interval into function k(x).

$k(x)=e^{sinx}$

$k(0)=e^{0}=1$

$k(\frac{\pi}{2})=e^{sin(\frac{\pi}{2})}=e^{1}=e$

$k(\frac{3\pi}{2})=e^{sin\frac{3\pi}{2}}=e^{-1}\approx 0.36$

$k(2\pi)=e^{sin(\pi)}=e^{0}=1$

Step 4: When we compare the above values, we got that 0.36 is the smallest and e is the greatest among them.

Therefore,

Absolute maximum value of k(x) is ‘e’ at $\frac{\pi}{2}$

Absolute minimum value of k(x) is $\frac{1}{e}$ at $\frac{3\pi}{2}$

Example 2: Find absolute maximum and minimum value of given function on given interval.

$g(x)=sin^2(x)+cosx;\left[\frac{2\pi}{3},\frac{3\pi}{2}\right]$

Points of Inflection

Points of inflection are those where a function changes its concavity. These are the points where a function’s graph changes from concave up to concave down or viceversa.

How to get points of inflection?

To get points of inflection, we find second derivative of given function, set it equal to 0 and solve for x.

What are possible points of inflection (PPOI) ?

PPOI are the x values where the second derivative of function is 0 or it doesn’t exist.

Are all PPOI are actually points of inflection?

Are all points where second derivative is 0 or doesn’t exist, points of inflection?

No, it is not necessary that all those points where second derivative is 0 or undefined, are points of inflection. Sometimes concavity doesn’t change at these points when we test the intervals. For example, for a function $y=x^{4}$ , second derivative is 0 at x=0 but still concavity doesn’t change at this point. So x=0 is not a point of inflection though second derivative is 0 here.

Find intervals of concave up and down from a derivative graph.

Find increasing/decreasing intervals, relative extrema , intervals of concavity and points of inflection from a given derivative graph

Critical numbers, intervals of increase and decrease, Relative extrema

February 1, 2025/by admin_ct

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

$f(x)=x^2 lnx$

Step 1: Find first derivative using product rule.

$f'(x)=(x^2)'lnx+x^2(lnx)'$

$f'(x)=2x lnx+x^2(\frac{1}{x})$

$f'(x)=2x lnx+x$

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

$2xlnx+x=0$

$x(2lnx+1)=0$

$x=0,2lnx+1=0$

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.

$\therefore 2lnx+1=0$

$lnx=\frac{-1}{2}$

$x=e^{\frac{-1}{2}}$ , rewriting the expression in terms of sqrt, we get

$x=\frac{1}{\sqrt{e}}$

So the only critical point is $x=\frac{1}{\sqrt{e}}$

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.
$g(x)=cos^2(x)$ on interval $[0,2\pi)$

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

Given function can be rewritten as,

$g(x)=(cosx)^2$

$g'(x)=2cosx*(cosx)'$

$g'(x)=2cosx(-sinx)=-2sinx cosx$

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

$-2sinxcosx=0$

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)

$x=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$

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 $(0,\frac{\pi}{2})$ . lets use pi/4 as testing point and plugin it into g'(x)

$g'(x)=-2sinx cosx$

$g'(x)=-2sin(\frac{\pi}{4})cos(\frac{\pi}{4})=-2(\frac{\sqrt{2}}{2})(\frac{\sqrt{2}}{2})=-1$

For interval $\left(\frac{\pi}{2},\pi\right)$ lets use 3pi/4 as testing point and plugin it into g'(x)

$g'(\frac{3\pi}{4})=-2sin(\frac{3\pi}{4})cos(\frac{3\pi}{4})=-2(\frac{\sqrt{2}}{2})(\frac{-\sqrt{2}}{2})=1$

For interval $(\pi,\frac{3\pi}{2})$ lets use 5pi/4 as testing point and plugin it into g'(x)

$g'(\frac{5\pi}{4})=-2sin(\frac{5\pi}{4})cos(\frac{5\pi}{4})=-2(\frac{-\sqrt{2}}{2})(\frac{-\sqrt{2}}{2})=-1$

For interval $(\frac{3\pi}{2},2\pi)$ lets use 7pi/4 as testing point and plugin it into g'(x)

$g'(\frac{7\pi}{4})=-2sin(\frac{7\pi}{4})cos(\frac{7\pi}{4})=-2(\frac{-\sqrt{2}}{2})(\frac{\sqrt{2}}{2})=1$

Next write all the results on a number line.

We get the following information from this sign chart.

Intervals of increase:

$(\frac{\pi}{2},\pi)\cup(\frac{3\pi}{2},2\pi)$

Intervals of decrease:

$(0,\frac{\pi}{2})\cup(\pi,\frac{3\pi}{2})$

Relative maxima:

$x=\pi$

Relative minima:

$x=\frac{\pi}{2},\frac{3\pi}{2}$

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 $f(x)=1-e^{g(x)}$ . 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.

$g(x)=\sqrt[3]{x^2-4}$
$g'(x)=(x-3)e^{x+2}$

Answers:

Increasing $(0,\infty)$ Decreasing $(-\infty,0)$
Increasing $(3,\infty)$ Decreasing $(-\infty,3)$

Mean value theorem

January 25, 2025/by admin_ct

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)

a) Is there a time t on interval $5\leq t\leq 10$ 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

$f(5)=45$

$f(10)=35$

That means this function h(t) take on every value from 45 to 35 on closed interval $[5,10]$ 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 $10\leqslant t\leqslant 40$ 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.

$h'(c)=\frac{h(b)-h(a)}{b-a}$

$h'(c)=\frac{h(40)-h(10)}{40-10}$

$h'(c)=\frac{125-35}{30}=3$

Hence proved that there exist a point ‘c’ on interval $10\leqslant t\leqslant 40$ such that $h'(c)=3$

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 $-4\leq x\leq 4$ is shown below. Does the mean value theorem applied on this interval guarantee a value ‘c’ for $-4<x<4$ 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 $-4<x<4$ 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.

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. $f(x)=x^2-5x+2;[-4,-2]$

b. $g(x)=sin(3x);[0,\pi]$

2. Find the equation of tangent line to the graph of f(x) = 2x+sinx+1 on the interval $\left[0,\pi\right]$ at the point which is the solution to the mean value theorem. Confirm your answer using graphic calculator.

Answers:

1. a) x= -3 b) $x=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$

2. $y-(\pi+2)=2(x-\frac{\pi}{2})$

Improper integrals

February 26, 2024/by admin_ct

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

December 25, 2023/by admin_ct

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 $[a, b]$ 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 $[a, b]$

Lets work on some examples:

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

December 23, 2023/by admin_ct

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.

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

Algebra of functions

October 5, 2023/by admin_ct

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.

Operation	Formula	Domain
Sum	(f+g)(x) = f(x) + g(x)	$A\cap B$
Difference	(f-g)(x) = f(x) – g(x)	$A\cap B$
Product	(fg)(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

September 17, 2023/by admin_ct

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:

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

Operations on Scientific Notations worksheet

August 25, 2023/by admin_ct

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 $\dpi{120} 8\times 10 ^{8} = 8\times 10^{2}\times 10^{6} = 800\times 10^{6}$