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

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

Figure $\PageIndex{1}$: Graphs of $y=\varphi(x)$ and $y=s(x)$

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 $\psi^{\prime} .$

Figure $\PageIndex{2}$: Graphs of $y=\psi(x)$ and $y=\psi^{\prime}(x)$.

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 $\mathbb{R}$.