
# 6.4: Discontinuities of Derivatives


## Theorem $$\PageIndex{1}$$: Intermediate Value Theorem for Derivatives

Suppose $$f$$ is differentiable on an open interval $$I, a, b \in I,$$ and $$a<b .$$ If $$\lambda \in \mathbb{R}$$ and either $$f^{\prime}(a)<\lambda<f^{\prime}(b)$$ or $$f^{\prime}(a)>\lambda>f^{\prime}(b),$$ then there exists $$c \in(a, b)$$ such that $$f^{\prime}(c)=\lambda$$.

Proof

Suppose $$f^{\prime}(a)<\lambda<f^{\prime}(b)$$ and define $$g: I \rightarrow \mathbb{R}$$ by $$g(x)=f(x)-\lambda x$$. Then $$g$$ is differentiable on $$I,$$ and so continuous on $$[a, b] .$$ Let $$c$$ be a point in $$[a, b]$$ at which $$g$$ attains its minimum value. Now

$g^{\prime}(a)=f^{\prime}(a)-\lambda<0,$

so there exists $$a<t<b$$ such that

$g(t)-g(a)<0.$

Thus $$c \neq a .$$ Similarly,

$g^{\prime}(b)=f^{\prime}(b)-\lambda>0,$

so there exists $$a<s<b$$ such that

$g(s)-g(b)<0.$

Thus $$c \neq b .$$ Hence $$c \in(a, b),$$ and so $$g^{\prime}(c)=0 .$$ Thus $$0=f^{\prime}(c)-\lambda,$$ and so $$f^{\prime}(c)=\lambda .$$ $$\quad$$ Q.E.D.

## Exercise $$\PageIndex{1}$$

Define $$g:(-1,1) \rightarrow \mathbb{R}$$ by

$g(x)=\left\{\begin{array}{cc}{-1,} & {\text { if }-1<x<0,} \\ {1,} & {\text { if } 0 \leq x<1.}\end{array}\right.$

Does there exist a function $$f:(-1,1) \rightarrow \mathbb{R}$$ such that $$f^{\prime}(x)=g(x)$$ for all $$x \in(-1,1) ?$$

## Exercise $$\PageIndex{2}$$

Suppose $$f$$ is differentiable on an open interval $$I .$$ Show that $$f^{\prime}$$ cannot have any simple discontinuities in $$I$$.

## Example $$\PageIndex{1}$$

Define $$\varphi:[0,1] \rightarrow \mathbb{R}$$ by $$\varphi(x)=x(2 x-1)(x-1) .$$ Define $$\rho: \mathbb{R} \rightarrow \mathbb{R}$$ by $$\rho(x)=6 x^{2}-6 x+1 .$$ Then

$\varphi(x)=2 x^{3}-3 x^{2}+x,$

so $$\varphi^{\prime}(x)=\rho(x)$$ for all $$x \in(0,1) .$$ Next define $$s: \mathbb{R} \rightarrow \mathbb{R}$$ by $$s(x)=\varphi(x-\lfloor x\rfloor)$$. See Figure $$\PageIndex{1}$$ for the graphs of $$\varphi$$ and $$s .$$ Then for any $$n \in \mathbb{Z}$$ and $$n<x<n+1$$,

$s^{\prime}(x)=\rho(x-n)=\rho(x-\lfloor x\rfloor).$

Moreover, if $$x$$ is an integer,

\begin{aligned} \lim _{h \rightarrow 0^{+}} \frac{s(x+h)-s(x)}{h} &=\lim _{h \rightarrow 0^{+}} \frac{\varphi(h)}{h} \\ &=\lim _{h \rightarrow 0^{+}} \frac{h(2 h-1)(h-1)}{h} \\ &=\lim _{h \rightarrow 0^{+}}(2 h-1)(h-1) \\ &=1 \end{aligned}

and

\begin{aligned} \lim _{h \rightarrow 0^{-}} \frac{s(x+h)-s(x)}{h} &=\lim _{h \rightarrow 0^{-}} \frac{\varphi(h+1)}{h} \\ &=\lim _{h \rightarrow 0^{-}} \frac{(h+1)(2 h+1) h}{h} \\ &=\lim _{h \rightarrow 0^{-}}(h+1)(2 h+1) \\ &=1. \end{aligned}

Thus $$s^{\prime}(x)=1=\rho(x-\lfloor x\rfloor)$$ when $$x$$ is an integer, and so $$s^{\prime}(x)=\rho(x-\lfloor x\rfloor)$$ for all $$x \in \mathbb{R} .$$

Now $$\rho(x)=0$$ if and only if $$x=\frac{3-\sqrt{3}}{6}$$ or $$x=\frac{3+\sqrt{3}}{6} .$$ since $$\varphi(0)=0$$, $$\varphi\left(\frac{3-\sqrt{3}}{6}\right)=\frac{1}{6 \sqrt{3}}, \varphi\left(\frac{3+\sqrt{3}}{6}\right)=-\frac{1}{6 \sqrt{3}},$$ and $$\varphi(1)=0,$$ we see that $$\varphi$$ attains a

maximum value of $$\frac{1}{6 \sqrt{3}}$$ and a minimum value of $$-\frac{1}{6 \sqrt{3}} .$$ Hence for any $$n \in \mathbb{Z}$$,

$s((n, n+1))=\left[-\frac{1}{6 \sqrt{3}}, \frac{1}{6 \sqrt{3}}\right].$

Also, $$\rho^{\prime}(x)=12 x-6,$$ so $$\rho^{\prime}(x)=0$$ if and only if $$x=\frac{1}{2} .$$ since $$\rho(0)=1$$, $$\rho\left(\frac{1}{2}\right)=-\frac{1}{2},$$ and $$\rho(1)=1,$$ we see that $$\rho$$ attains a maximum value of 1 and a v,

$s^{\prime}((n, n+1))=\left[-\frac{1}{2}, 1\right].$

It follows from the preceding, in the same manner as the result in Example $$5.1 .7,$$ that neither the function $$\sigma(x)=s\left(\frac{1}{x}\right)$$ nor the function $$g(x)=s^{\prime}\left(\frac{1}{x}\right)$$ has a limit as $$x$$ approaches $$0 .$$

Finally, define $$\psi: \mathbb{R} \rightarrow \mathbb{R}$$ by

$\psi(x)=\left\{\begin{array}{ll}{x^{2} s\left(\frac{1}{x}\right),} & {\text { if } x \neq 0,} \\ {0,} & {\text { if } x=0.}\end{array}\right.$

For $$x \neq 0,$$ we have

$\psi^{\prime}(x)=x^{2} s^{\prime}\left(\frac{1}{x}\right)\left(-\frac{1}{x^{2}}\right)+2 x s\left(\frac{1}{x}\right)=-s^{\prime}\left(\frac{1}{x}\right)+2 x s\left(\frac{1}{x}\right).$

At $$0,$$ we have

\begin{aligned} \psi^{\prime}(0) &=\lim _{h \rightarrow 0} \frac{\psi(0+h)-\psi(0)}{h} \\ &=\lim _{h \rightarrow 0} \frac{h^{2} s\left(\frac{1}{h}\right)}{h} \\ &=\lim _{h \rightarrow 0} h s\left(\frac{1}{h}\right) \\ &=0, \end{aligned}

where the final limit follows from the squeeze theorem and the fact that $$s$$ is bounded. Hence we see that $$\psi$$ is continuous on $$\mathbb{R}$$ and differentiable on $$\mathbb{R},$$ but $$\psi^{\prime}$$ is not continuous since $$\psi^{\prime}(x)$$ does not have a limit as $$x$$ approaches $$0 .$$ See Figure $$\PageIndex{1}$$ for the graphs of $$\psi$$ and $$\psi^{\prime} .$$

## Exercise $$\PageIndex{3}$$

Let $$s$$ be as above and define $$g: \mathbb{R} \rightarrow \mathbb{R}$$ by

$g(x)=\left\{\begin{array}{ll}{x^{4} s\left(\frac{1}{x}\right),} & {\text { if } x \neq 0} \\ {0,} & {\text { if } x=0.}\end{array}\right.$

Show that $$g$$ is differentiable on $$\mathbb{R}$$ and that $$g^{\prime}$$ is continuous on $$\mathbb{R}$$.