CLP-1 Differential Calculus

Subsection 2.13.4 Be Careful with Hypotheses

The mean value theorem has hypotheses — \(f(x)\) has to be continuous for \(a\le x\le b\) and has to be differentiable for \(a \lt x \lt b\text{.}\) If either hypothesis is violated, the conclusion of the mean value theorem can fail. That is, the curve \(y=f(x)\) need not have a tangent line at some \(x=c\) between \(a\) and \(b\) whose slope, \(f'(c)\text{,}\) is the same as the slope, \(\frac{f(b)-f(a)}{b-a}\text{,}\) of the secant joining the points \(\big(a\,,f(a)\big)\) and \(\big(b\,,f(b)\big)\) on the curve. If \(f'(x)\) fails to exist for even a single value of \(x\) between \(a\) and \(b\text{,}\) all bets are off. The following two examples illustrate this.

For the first “bad” example, \(a=0\text{,}\) \(b=2\) and

\(f(x) = \begin{cases} 0 & \text{if }x \le 1 \\ 1 & \text{if }x \gt 1 \end{cases}\)

For this example, \(f'(x)=0\) at every \(x\) where it is defined. That is, at every \(x\ne 1\text{.}\) But the slope of the secant joining \(\big(a\,,f(a)\big)=(0,0)\) and \(\big(b\,,f(b)\big)=(2,1)\) is \(\frac{1}{2}\text{.}\)

For the second “bad” example, \(a=-1\text{,}\) \(b=1\) and \(f(x)=|x|\text{.}\) For this function

\(f'(x) = \begin{cases} -1 & \text{if }x \lt 0 \\ \text{undefined} & \text{if }x=0 \\ 1 & \text{if }x \gt 0 \end{cases}\)

For this example, \(f'(x)=\pm 1\) at every \(x\) where it is defined. That is, at every \(x\ne 0\text{.}\) But the slope of the secant joining \(\big(a\,,f(a)\big)=(-1,1)\) and \(\big(b\,,f(b)\big)=(1,1)\) is \(0\text{.}\)

Here is one “good” example, where the hypotheses of the mean value theorem are satisfied. Let \(f(x)=x^2\text{.}\) Then \(f'(x)=2x\text{.}\) For any \(a \lt b\text{,}\)

\begin{gather*} \frac{f(b)-f(a)}{b-a}=\frac{b^2-a^2}{b-a}=b+a \end{gather*}

So \(f'(c)=2c\) is exactly \(\frac{f(b)-f(a)}{b-a}\) when \(c=\frac{a+b}{2}\text{,}\) which, in this example, happens to be exactly half way between \(x=a\) and \(x=b\text{.}\)

Recall from Section 2.3 that if \(f'(c)>0\text{,}\) then \(f(x)\) is increasing at \(x=c\text{.}\) A simple consequence of the mean value theorem is that if you know the sign of \(f'(c)\) for all \(c\)'s between \(a\) and \(b\text{,}\) with \(b \gt a\text{,}\) then \(f(b)-f(a) = f'(c) (b-a)\) must have the same sign.

It is not hard to see why the above is true:

• Say \(f'(x)=0\) at every point in the interval \([A,B]\text{.}\) Now pick any \(a,b \in [A,B]\) with \(a \lt b\text{.}\) Then the MVT tells us that there is \(c \in (a,b)\) so that
  \begin{align*} f'(c) &= \frac{f(b)-f(a)}{b-a} \end{align*}
  If \(f(b) \neq f(a)\) then we must have that \(f'(c) \neq 0\) — contradicting what we are told about \(f'(x)\text{.}\) Thus we must have that \(f(b)=f(a)\text{.}\)
• Similarly, say \(f'(x) \geq 0\) at every point in the interval \([A,B]\text{.}\) Now pick any \(a,b \in [A,B]\) with \(a \lt b\text{.}\) Then the MVT tells us that there is \(c \in (a,b)\) so that
  \begin{align*} f'(c) &= \frac{f(b)-f(a)}{b-a} \end{align*}
  Since \(b \gt a\text{,}\) the denominator is positive. Now if \(f(b) \lt f(a)\) the numerator would be negative, making the right-hand side negative, and contradicting what we are told about \(f'(x)\text{.}\) Hence we must have \(f(b) \ge f(a)\text{.}\)

A nice corollary of the above corollary is the following:

We can prove this by setting \(h(x)=f(x)-g(x)\text{.}\) Then \(h'(x)=0\) and so the previous corollary tells us that \(h(x)\) is constant.