# 4.4 The mean value theorem  (Page 4/7)

 Page 4 / 7
$\frac{f\left(b\right)-f\left(a\right)}{b-a}\ne 0.$

Since $f$ is a differentiable function, by the Mean Value Theorem, there exists $c\in \left(a,b\right)$ such that

${f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

Therefore, there exists $c\in I$ such that ${f}^{\prime }\left(c\right)\ne 0,$ which contradicts the assumption that ${f}^{\prime }\left(x\right)=0$ for all $x\in I.$

From [link] , it follows that if two functions have the same derivative, they differ by, at most, a constant.

## Corollary 2: constant difference theorem

If $f$ and $g$ are differentiable over an interval $I$ and ${f}^{\prime }\left(x\right)={g}^{\prime }\left(x\right)$ for all $x\in I,$ then $f\left(x\right)=g\left(x\right)+C$ for some constant $C.$

## Proof

Let $h\left(x\right)=f\left(x\right)-g\left(x\right).$ Then, ${h}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)-{g}^{\prime }\left(x\right)=0$ for all $x\in I.$ By Corollary 1, there is a constant $C$ such that $h\left(x\right)=C$ for all $x\in I.$ Therefore, $f\left(x\right)=g\left(x\right)+C$ for all $x\in I.$

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function $f$ is increasing over $I$ if $f\left({x}_{1}\right) whenever ${x}_{1}<{x}_{2},$ whereas $f$ is decreasing over $I$ if $f{\left(x\right)}_{1}>f\left({x}_{2}\right)$ whenever ${x}_{1}<{x}_{2}.$ Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing ( [link] ). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph.

This fact is important because it means that for a given function $f,$ if there exists a function $F$ such that ${F}^{\prime }\left(x\right)=f\left(x\right);$ then, the only other functions that have a derivative equal to $f$ are $F\left(x\right)+C$ for some constant $C.$ We discuss this result in more detail later in the chapter.

## Corollary 3: increasing and decreasing functions

Let $f$ be continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right).$

1. If ${f}^{\prime }\left(x\right)>0$ for all $x\in \left(a,b\right),$ then $f$ is an increasing function over $\left[a,b\right].$
2. If ${f}^{\prime }\left(x\right)<0$ for all $x\in \left(a,b\right),$ then $f$ is a decreasing function over $\left[a,b\right].$

## Proof

We will prove i.; the proof of ii. is similar. Suppose $f$ is not an increasing function on $I.$ Then there exist $a$ and $b$ in $I$ such that $a but $f\left(a\right)\ge f\left(b\right).$ Since $f$ is a differentiable function over $I,$ by the Mean Value Theorem there exists $c\in \left(a,b\right)$ such that

${f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

Since $f\left(a\right)\ge f\left(b\right),$ we know that $f\left(b\right)-f\left(a\right)\le 0.$ Also, $a tells us that $b-a>0.$ We conclude that

${f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\le 0.$

However, ${f}^{\prime }\left(x\right)>0$ for all $x\in I.$ This is a contradiction, and therefore $f$ must be an increasing function over $I.$

## Key concepts

• If $f$ is continuous over $\left[a,b\right]$ and differentiable over $\left(a,b\right)$ and $f\left(a\right)=0=f\left(b\right),$ then there exists a point $c\in \left(a,b\right)$ such that ${f}^{\prime }\left(c\right)=0.$ This is Rolle’s theorem.
• If $f$ is continuous over $\left[a,b\right]$ and differentiable over $\left(a,b\right),$ then there exists a point $c\in \left(a,b\right)$ such that
$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

This is the Mean Value Theorem.
• If $f\prime \left(x\right)=0$ over an interval $I,$ then $f$ is constant over $I.$
• If two differentiable functions $f$ and $g$ satisfy ${f}^{\prime }\left(x\right)={g}^{\prime }\left(x\right)$ over $I,$ then $f\left(x\right)=g\left(x\right)+C$ for some constant $C.$
• If ${f}^{\prime }\left(x\right)>0$ over an interval $I,$ then $f$ is increasing over $I.$ If ${f}^{\prime }\left(x\right)<0$ over $I,$ then $f$ is decreasing over $I.$

#### Questions & Answers

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