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 \(\mathbb{R}\).