4.4 The mean value theorem

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

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 $x\in I.$

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

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 $h(x)=C$ for all $x\in I.$ Therefore, $f(x)=g(x)+C$ for all $x\in I.$

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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 $f\left(x_1\right)<f\left(x_2\right)$ 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 $F(x)+C$ for some constant $C.$ We discuss this result in more detail later in the chapter.

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 $a<b,$ 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

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

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 $I.$

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Key concepts

• If $f$ is continuous over $[a,b]$ 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 $[a,b]$ 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 $f\prime (x)=0$ over an interval $I,$ then $f$ is constant over $I.$
• If two differentiable functions $f$ and $g$ satisfy $f^{\prime }(x)=g^{\prime }(x)$ 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 $f^{\prime }(x)<0$ over $I,$ then $f$ is decreasing over $I.$