# 4.4 The mean value theorem

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• Explain the meaning of Rolle’s theorem.
• Describe the significance of the Mean Value Theorem.
• State three important consequences of the Mean Value Theorem.

The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem.

## Rolle’s theorem

Informally, Rolle’s theorem states that if the outputs of a differentiable function $f$ are equal at the endpoints of an interval, then there must be an interior point $c$ where $f\prime \left(c\right)=0.$ [link] illustrates this theorem.

## Rolle’s theorem

Let $f$ be a continuous function over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right)$ such that $f\left(a\right)=f\left(b\right).$ There then exists at least one $c\in \left(a,b\right)$ such that $f\prime \left(c\right)=0.$

## Proof

Let $k=f\left(a\right)=f\left(b\right).$ We consider three cases:

1. $f\left(x\right)=k$ for all $x\in \left(a,b\right).$
2. There exists $x\in \left(a,b\right)$ such that $f\left(x\right)>k.$
3. There exists $x\in \left(a,b\right)$ such that $f\left(x\right)

Case 1: If $f\left(x\right)=0$ for all $x\in \left(a,b\right),$ then $f\prime \left(x\right)=0$ for all $x\in \left(a,b\right).$

Case 2: Since $f$ is a continuous function over the closed, bounded interval $\left[a,b\right],$ by the extreme value theorem, it has an absolute maximum. Also, since there is a point $x\in \left(a,b\right)$ such that $f\left(x\right)>k,$ the absolute maximum is greater than $k.$ Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point $c\in \left(a,b\right).$ Because $f$ has a maximum at an interior point $c,$ and $f$ is differentiable at $c,$ by Fermat’s theorem, $f\prime \left(c\right)=0.$

Case 3: The case when there exists a point $x\in \left(a,b\right)$ such that $f\left(x\right) is analogous to case 2, with maximum replaced by minimum.

An important point about Rolle’s theorem is that the differentiability of the function $f$ is critical. If $f$ is not differentiable, even at a single point, the result may not hold. For example, the function $f\left(x\right)=|x|-1$ is continuous over $\left[-1,1\right]$ and $f\left(-1\right)=0=f\left(1\right),$ but $f\prime \left(c\right)\ne 0$ for any $c\in \left(-1,1\right)$ as shown in the following figure.

Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points $c$ where $f\prime \left(c\right)=0.$

## Using rolle’s theorem

For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values $c$ in the given interval where $f\prime \left(c\right)=0.$

1. $f\left(x\right)={x}^{2}+2x$ over $\left[-2,0\right]$
2. $f\left(x\right)={x}^{3}-4x$ over $\left[-2,2\right]$
1. Since $f$ is a polynomial, it is continuous and differentiable everywhere. In addition, $f\left(-2\right)=0=f\left(0\right).$ Therefore, $f$ satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value $c\in \left(-2,0\right)$ such that $f\prime \left(c\right)=0.$ Since $f\prime \left(x\right)=2x+2=2\left(x+1\right),$ we see that $f\prime \left(c\right)=2\left(c+1\right)=0$ implies $c=-1$ as shown in the following graph.
2. As in part a. $f$ is a polynomial and therefore is continuous and differentiable everywhere. Also, $f\left(-2\right)=0=f\left(2\right).$ That said, $f$ satisfies the criteria of Rolle’s theorem. Differentiating, we find that $f\prime \left(x\right)=3{x}^{2}-4.$ Therefore, $f\prime \left(c\right)=0$ when $x=\text{±}\frac{2}{\sqrt{3}}.$ Both points are in the interval $\left[-2,2\right],$ and, therefore, both points satisfy the conclusion of Rolle’s theorem as shown in the following graph.

#### Questions & Answers

what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
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first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
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a variable that does not depend on another.
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solve number one step by step
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel
how to prove tan^2 x=csc^2 x tan^2 x-1?
divide by tan^2 x giving 1=csc^2 x -1/tan^2 x, rewrite as: 1=1/sin^2 x -cos^2 x/sin^2 x, multiply by sin^2 x giving: sin^2 x=1-cos^2x. rewrite as the familiar sin^2 x + cos^2x=1 QED
Barnabas
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sin x - sin x cos^2 x sin x (1-cos^2 x) note the identity:sin^2 x + cos^2 x = 1 thus, sin^2 x = 1 - cos^2 x now substitute this into the above: sin x (sin^2 x), now multiply, yielding: sin^3 x Q.E.D.
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navin
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an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
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a term/algebraic expression raised to a non-negative integer power and a multiple of co-efficient,,,,,, T^n where n is a non-negative,,,,, 4x^2
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find volume of solid about y axis and y=x^3, x=0,y=1
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Is a rule used to find a derivative. For example the derivative of y(x)= a(x)^n is y'(x)= a*n*x^n-1.
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