Proof. Suppose \(f\) has a local maximum at \(c\): there is \(R>0\) such that \((c - R, c+ R)\subseteq S\) and \(f(x) \leq f(c)\) for \(c- R < x < c + R\). If \(c - R < x < c\) then \(\frac {f(x) - f(c)}{x - c}\geq 0\), so \begin {equation} \label {eq:extleft} f'(c) = \lim _{x\rightarrow c^-} \frac {f(x) - f(c)}{x - c}\geq 0 . \end {equation} Similarly, if \(c < x < c + R\) then \(\frac {f(x) - f(c) }{x - c} \leq 0\), so that \begin {equation} \label {eq:extright} f'(c) = \lim _{x\rightarrow c^+} \frac {f(x) - f(c)}{x - c}\leq 0 . \end {equation} Since \(f'(c)\) exists, it satisfies both (4.5) and (4.6), so \(f'(c) = 0\). If \(f\) has a local minimum at \(c\), then \(-f\) has a local maximum there, and \(-f'(c) = 0\). □

Proof. Define \(F: I\longrightarrow \mathbb {R}\) by \(F(x) = f(x) - Kx\). Then \(F'(a) = f'(a) - K < 0\) and \(F'(b) = f'(b) - K > 0\). Since \(\frac {F(a + h) - F(a)}{h} \longrightarrow F'(a)<0\) as \(h \longrightarrow 0\), we have \(F(a+h) < F(a)\) for all sufficiently small \(h>0\); so there is \(t_1\in (a,b)\) with \(F(t_1)<F(a)\). Similarly, since \(\frac {F(b + h) - F(b)}{h} \longrightarrow F'(b)>0\), we have \(F(b+h) < F(b)\) for all sufficiently small \(h<0\); so there is \(t_2\in (a,b)\) with \(F(t_2)<F(b)\).

The function \(F\) is differentiable, hence continuous, on \([a,b]\), so it attains its minimum on \([a,b]\) at some point \(c\). Since \(F(t_1)<F(a)\) and \(F(t_2)<F(b)\), the minimum is not attained at \(a\) or \(b\), so \(a<c<b\) and \(F\) has a local minimum at \(c\). By the previous theorem \(0 = F'(c) = f'(c) - K\), that is, \(f'(c) = K\). For the case \(f'(a) > K > f'(b)\), apply this to \(-f\) and \(-K\). □

Solution. Here \(f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)\). So \(f' > 0\) on \((-\infty ,1)\), \(f'<0\) on \((1,3)\) and \(f'>0\) on \((3,\infty )\). By the Mean Value Theorem, \(f\) increases, then decreases, then increases: there is a local maximum at \(x=1\) with \(f(1) = 5\) and a local minimum at \(x=3\) with \(f(3) = 1\).

On \([0,5]\), the global extrema occur either at the end points or at interior points where \(f' = 0\). Comparing \[f(0) = 1,\qquad f(1) = 5,\qquad f(3) = 1,\qquad f(5) = 21,\] the global maximum is \(21\), attained at the end point \(5\), and the global minimum is \(1\), attained at \(0\) and at \(3\). The local maximum at \(1\) is not a global maximum.

Solution. For \(f(x) = x^3\), \(f'(0) = 0\), but \(f\) is strictly increasing, so it has no local maximum or minimum at \(0\). A zero of \(f'\) is only a candidate for an extremum.

For \(f(x) = x\) on \([0,1]\), the maximum is at the end point \(1\), where \(f'(1) = 1\neq 0\). End points must always be checked separately.

Solution. If \(\operatorname {sgn} = f'\) for some differentiable \(f\), then \(f'(-1) = -1 < \frac {1}{2} < 1 = f'(1)\), and by the Intermediate Value Theorem for Derivatives there would be a \(c\) with \(f'(c) = \frac {1}{2}\). But \(\operatorname {sgn}\) takes only the values \(-1\), \(0\), \(1\). So no such \(f\) exists. In general, a derivative cannot have a jump discontinuity, even though it need not be continuous (compare \(g'\) in the example with \(x^2\sin \frac {1}{x}\)).