The proofs of (2) and (3) are nearly identical. We prove (2). Assume \(f'(x)> 0\) for all interior points \(x\in I\text{.}\) Suppose \(a,b\in I\) satisfy \(a< b\text{.}\) Since the function \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\text{,}\) the mean value theorem implies
Before putting TheoremΒ 22.2 to good analytic work on some example functions, we first illustrate that we cannot apply the theorem if the underlying set is not an interval.
Let \(f(x)=\frac{1}{x}\) with domain \(D=\R-\{0\}\text{.}\) Observe that \(f'(x)=-1/x^2\) is negative for all \(x\in D\text{.}\) However it is not true that \(f\) is decreasing on its entire domain \(D\text{:}\) indeed, we have \(-1< 1\text{,}\) and yet \(-1=f(-1)\not> f(1)=1\text{.}\)
a union of two intervals. We are permitted to apply TheoremΒ 22.2 to \(I_1\) and \(I_2\) separately to conclude that \(f\) is decreasing on each of these intervals, but this does not imply that it is increasing on their union.
According to TheoremΒ 22.2 determining the sets on which a function \(f\) is increasing/decreasing essentially amounts to solving the inequalities \(f'(x)> 0\) and \(f'(x)< 0\text{.}\) This is most efficiently done using a sign diagram of \(f'\text{,}\) as illustrated in the next example and defined more precisely in ProcedureΒ 22.5.
Since \(f'\) is continuous, and since \(0,1,-1\) are the only roots of \(f'\text{,}\) the intermediate value theorem implies that \(f'\) is either always positive or always negative on each of the resulting intervals \((-\infty,-1),(-1,0),(0,1),(1,\infty)\text{.}\) This means we can determine the sign of \(f'\) on each interval by evaluating it at one test point. It is easy to see that
From the sign diagram we conclude that \(f\) is increasing on the intervals \((-\infty, -1]\) and \([1,\infty)\text{,}\) and decreasing on the interval \([-1,1]\text{.}\)
If \(f'\) is nonzero near \(c\) and the sign does not change at \(c\text{,}\) then \(f(c)\) is neither a local maximum value nor a local minimum value of \(f\text{.}\)
Procedure22.7.Classify critical points: first derivative test.
Let \(f\) be continuous on the set \(D\text{.}\) To find and classify all critical points of \(f\) as corresponding to a local maximum value of \(f\text{,}\) a local minimum value of \(f\text{,}\) or neither, proceed as follows.
For each critical point \(a\) of \(f\text{,}\) use TheoremΒ 22.6 and your sign diagram to determine whether \(f(a)\) is a local maximum value of \(f\text{,}\) local minimum value of \(f\text{,}\) or neither.
Let \(f(x)=3x^5-5x^3\text{.}\) Find all critical points of \(f\text{,}\) and for each critical point \(a\) classify \(f(a)\) as a local maximum value of \(f\text{,}\) a local minimum value of \(f\text{,}\) or neither.
In the course of the sign diagram from ExampleΒ 22.4 we saw that \(x=-1,0,1\) were the critical points of \(f\text{.}\) Since the sign of \(f'\) switches from positive to negative at \(-1\text{,}\)\(f(-1)\) is a local maximum value of \(f\text{;}\) since the sign of \(f'\) switches from negative to positive at \(1\text{,}\)\(f(1)\) is a local minimum value of \(f\text{.}\) Since \(f'(x)\ne 0\) for \(x\) close to \(0\) and since the sign of \(f'\) does not switch at \(x=0\text{,}\) we see that \(f(0)\) is neither a local maximum nor a local minimum value.
Example22.9.Monotonicity and critical points: trig.
Let \(f(x)=2\cos x+\sqrt{2}\cos^2 x\) and let \(I=[\pi/2, 3\pi/2]\text{.}\)
Find all critical points of \(f\text{,}\) and for each critical point \(a\) classify \(f(a)\) as a local maximum value of \(f\text{,}\) a local minimum value of \(f\text{,}\) or neither.
Since \(f\) is differentiable everywhere, critical points of \(f\) are solutions to \(f'(x)=0\text{.}\) We thus solve:
\begin{align*}
f'(x) \amp =0\\
-2\sin x(1+\sqrt{2}\cos x) \amp = 0\\
\sin x =0 \amp\text{ or } 1+\sqrt{2}\cos x=0 \\
\sin x =0 \amp \text{ or } \cos x=-\frac{1}{\sqrt{2}} \text{.}
\end{align*}
We have \(\sin x=0\) if and only if \(x=\pi n\) for some integer \(n\in \Z\text{.}\) Only one of these solutions lies in \(I=[\pi/2,3\pi /2]\text{:}\) namely, \(x=\pi\text{.}\) Next, observing that \(1/\sqrt{2}=\sqrt{2}/2\text{,}\) and recalling facts about the unit circle, we see that \(x\) solves \(\cos x=-1/\sqrt{2}=-\sqrt{2}/2\) if and only if
\begin{equation*}
x=\frac{3\pi }{4}+2\pi n \text{ or } x=\frac{5\pi }{4}+2\pi n
\end{equation*}
for some integer \(n\in \Z\text{.}\) There are exactly two such solutions lying in \([\pi/2, 3\pi/2]\text{:}\) namely, \(x=3\pi/4\) and \(x=5\pi /4\text{.}\) In conclusion, the critical points of \(f\) are \(3\pi/4, \pi, 5\pi/4\text{.}\)
It follows that \(f(3\pi/4)\) is a local minimum value, \(f(\pi)\) is a local maximum value, and \(f(5\pi/4)\) is a local minimum value of \(f\text{.}\)
Furthermore, we see that within the interval \([\pi,3\pi/2]\) the function \(f\) is increasing on \([\pi,3\pi/4]\) and \([\pi, 5\pi /4]\text{,}\) and decreasing on \([3\pi/4, \pi]\) and \([5\pi/4, 3\pi /2]\text{.}\)
Example22.11.Monotonicity and critical points: radical.
Let \(f(x)=x\, \sqrt{7-x^2}\text{.}\) Find all critical points of \(f\text{,}\) and for each critical point \(a\) classify \(f(a)\) as a local maximum value of \(f\text{,}\) local minimum value of \(f\text{,}\) or neither.
Following ProcedureΒ 22.7, we first determine the critical points of \(f\text{.}\) Note first that in this case \(f\) is not differentiable at \(\pm \sqrt{7}\text{,}\) making these critical points. Next, for all \(x\ne \pm \sqrt{7}\text{,}\) we compute \(f'(x)\) using the product and chain rules:
It follows that we have \(f'(x)=0\) if and only if \(7-2x^2=0\text{,}\) or equivalently, \(x=\pm \sqrt{7/2}\text{.}\) We conclude that there are in total four critical points: \(x=\pm \sqrt{7}\) (where \(f\) is not differentiable), and \(x=\pm\sqrt{7/2}\text{.}\)
We conclude that \(f(-\sqrt{7})\) and \(f(\sqrt{7/2})\) are local maximum values, and \(f(-\sqrt{7/2}\) and \(f(\sqrt{7}\) are local minimum values. These conclusions are verified by the actual graph of \(f\text{.}\)