7.2 Important Theorems about Derivatives

The following theorems about the properties of differential functions are among the most important results in real analysis for applied mathematics and physics. The first theorem we will look at is Rolle’s theorem, which essentially states that if a function is differentiable and has the same value at two points, then the derivative must be zero somewhere between those two points.

7.2.1 Rolle’s Theorem

Theorem 7.4 (Rolle’s Theorem)

Let $a,b\in\mathbb{R}$ with $a<b$ , let $f:[a,b]\to\mathbb{R}$ be a continuous function which is differentiable on $(a,b)$ with $f(a)=f(b)$ . Then there exists $\theta\in(a,b)$ such that $f^{\prime}(\theta)=0$ .

Proof.

Firstly, we will prove that the derivative of a function is zero at the maximum or minimum points of a function. This is a fact that we are used to from school, and we will need it when proving this theorem.

Lemma

Let $I\subseteq\mathbb{R}$ be an open interval and $f:I\to\mathbb{R}$ be a differentiable function.

(i)

If $f$ attains its maximum at some $c\in I$ , then $f^{\prime}(c)=0$ .
(ii)

If $f$ attains its minimum at some $c\in I$ , then $f^{\prime}(c)=0$ .

Proof.

(i)

Since $f$ has reached its maximum, $f(x)\leq f(c)\ \forall x\in I$ . Hence

$f(x)-f(c)\leq 0\ \forall x\in I.$ (7.26)

Now, if $x<c$ , then $x-c<0$ and

$\frac{f(x)-f(c)}{x-c}\geq 0,\ \text{therefore}\ \lim_{x\to c^{-}}\frac{f(x)-f(% c)}{x-c}\geq 0.$ (7.27)

On the other hand, if $x>c$ , then $x-c>0$ and

$\frac{f(x)-f(c)}{x-c}\leq 0,\ \text{therefore}\ \lim_{x\to c^{+}}\frac{f(x)-f(% c)}{x-c}\leq 0.$ (7.28)

Since $f$ is differentiable, $f^{\prime}(c)$ is defined and so the left and right limits at $c$ must be equal, which can only be true if

$f^{\prime}(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=0.$ (7.29)
(ii)

follows similarly.

∎

Now there are three cases to consider. The first is the case where the function is a horizontal line between $a$ and $b$ , the second is the case where the function has a maximum between $a$ and $b$ , and the third is the case where there is a minimum.

(i)

Suppose $\forall x\in[a,b]$ , $f(x)=f(a)$ . Then $\forall\theta\in(a,b)$ , $f^{\prime}(\theta)=0$ .
(ii)

Suppose there exists some $y\in[a,b]$ with $f(y)>f(a)=f(b)$ . Since $f$ is continuous on $[a,b]$ , the Extreme Value Theorem (6.4) says that $f$ attains a maximum at some $\theta\in[a,b]$ , therefore

$f(x)\leq f(\theta)\ \forall x\in[a,b].$ (7.30)

However since $f(y)>f(a)=f(b)$ , $\theta$ must be at least $y$ , i.e. $\theta\neq a$ , $\theta\neq b$ . So $\theta\in(a,b)$ and hence by the part (i)) of the lemma above, $f^{\prime}(\theta)=0$ .
(iii)

Suppose there exists some $y\in[a,b]$ with $f(y)<f(a)=f(b)$ . Then this case follows similarly to (ii) in that $f$ must attain a minimum in $(a,b)$ .

∎

7.2.2 The Mean Value Theorem

The next theorem is a generalisation of Rolle’s theorem. It states that between two points of a function, there must be a least one point where the tangent to the function is parallel to a straight line between the two points.

Theorem 7.5 (Mean Value Theorem)

Let $a,b\in\mathbb{R}$ with $a<b$ and let $f:[a,b]\to\mathbb{R}$ be a continuous function which is differentiable on $(a,b)$ . Then $\exists\theta\in(a,b)$ such that

f^{\prime}(\theta)=\frac{f(b)-f(a)}{b-a}.

(7.31)

Proof.

This theorem is very easy to prove using Rolle’s theorem (7.4). Let $g(x)=f(x)-\lambda x$ , where $\lambda$ is chosen so that $g(a)=g(b)$ . So

$\displaystyle f(a)-\lambda a$	$\displaystyle=f(b)-\lambda b$	(7.32)
$\displaystyle\lambda(b-a)$	$\displaystyle=f(b)-f(a)$	(7.33)
$\displaystyle\lambda$	$\displaystyle=\frac{f(b)-f(a)}{b-a}.$	(7.34)

Then by Rolle’s theorem, there exists $\theta\in(a,b)$ such that $g^{\prime}(\theta)=0$ . By theorem 7.2,

	$\displaystyle g^{\prime}(\theta)$	$\displaystyle=f^{\prime}(\theta)-\lambda=0$		(7.35)
	$\displaystyle\implies f^{\prime}(\theta)$	$\displaystyle=\lambda=\frac{f(b)-f(a)}{b-a}.$		(7.36)

∎

Another way of writing the Mean Value theorem is that there exists $\theta\in(a,b)$ so that $f(b)=f(a)+(b-a)f^{\prime}(\theta)$ . This is another way of expressing what it means for a function to be differentiable. If we replace $b$ with some point $x$ in the interior of the domain of $f$ then we get

f(x)=f(a)+f^{\prime}(\theta)(x-a).

(7.37)

This is a linear function, so we say that if a function is differentiable at a point then it has a good linear approximation at that point. This fact could be used to calculate the value of the function at lots of points if we know the value of the function at a few points and we know information about the derivative. We will touch on this in the last part of this chapter.

Example 7.5.

Suppose $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function with $f(0)=0,f(1)=1,f(2)=2$ . Show that $f^{\prime}(x)$ takes the values 0 and $\frac{1}{2}$ for some $x\in(0,2)$ .
Proof: Note that $f(1)=f(2)$ , so by Rolle’s theorem there exists some $\theta\in(1,2)\subseteq(0,2)$ with $f^{\prime}(\theta)=0$ . Also, by the Mean Value theorem there exists $\theta\in(0,2)$ such that

f^{\prime}(\theta)=\frac{f(2)-f(0)}{2-0}=\frac{1-0}{2-0}=\frac{1}{2}.

(7.38)