Although the domain of \(f\) is \(\R\text{,}\) observe that since \(f(x+2\pi)=f(x)\) for all \(x\in \R\text{,}\) we can obtain a graph of \(f\) by producing a careful sketch on the interval \([0,2\pi]\) and then “copying and pasting” this graph over all other intervals \([2\pi n, 2\pi(n+1)]\text{.}\) This means for the moment our analysis will be restricted to just the interval \([0,2\pi]\text{.}\) The \(y\)-intercept, as always, is \((0,f(0))=(0,0)\text{.}\) To find the \(x\)-intercepts in \([0,2\pi]\) we solve:
\begin{align*}
f(x) \amp =0\\
\sin x(1+\sin x) \amp = 0\\
\sin x= 0 \amp \text{ or } \sin x=-1\\
x \amp =0, \pi, 3\pi/2,2\pi\text{.}
\end{align*}
Next we compute
\begin{align*}
f'(x) \amp = \cos x(1+2\sin x)\\
f''(x) \amp = -\sin x+2\cos^2(x)-2\sin^2 x=2-\sin x-4\sin^2 x \text{.}
\end{align*}
From the factored form of \(f'\text{,}\) we see solve
\begin{align*}
f'(x) \amp = 0\\
\cos x=0 \amp \text{ or } \sin x=-\frac{1}{2}\\
x \amp =\frac{\pi}{2}, \frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\text{.}
\end{align*}
With a little work, choosing our test points carefully, we produce the following sign diagram for \(f'\) on the interval \([0,2\pi]\text{.}\)
The diagram clearly indicates the intervals of monotonicity within \([0,2\pi]\text{,}\) and indicates that \(f\) attains in alternating manner local maximum/minimum values at these inputs.
From this information, and the table of values
\begin{equation*}
\begin{array}{r|l}
x \amp f(x) \\
\hline
0 \amp 0 \\
\pi \amp 0 \\
\frac{3\pi}{2} \amp 0 \\
2\pi \amp 0 \\
\frac{\pi}{2} \amp 2 \\
\frac{7\pi}{6} \amp -\frac{1}{4} \\
\frac{11\pi}{6}\amp -\frac{1}{4}
\end{array}
\end{equation*}
we can obtain the following sketch of \(f\) on \([0,2\pi]\text{.}\)
As mentioned above, using the periodicity of \(f\text{,}\) we get a full sketch of \(f\) on \(\R\) by “cutting and pasting”.
Note that as a result of the periodicity of
\(f\text{,}\) the absolute maximum and minimum values of
\(f\) on the domain
\(\R\) are equal to the absolute maximum and minimum values of
\(f\) on the interval
\([0,2\pi]\text{.}\) From our table of values above, which includes evaluations at all critical points in
\([0,2\pi]\text{,}\) along with
Procedure 20.13, we see that
\(f(\pi/2)=2\) is the absolute maximum value of
\(f\) on
\(\R\text{,}\) and
\(f(7\pi/6)=f(11\pi/6)=-1/4\) is the absolute minimum value.
What about the concavity of \(f\) and inflection points? The second derivative
\begin{equation*}
f''(x)=-4\sin ^2(x)-\sin x+2
\end{equation*}
is somewhat complicated, making a precise sign diagram for \(f''\) difficult to produce. Our graph of \(f\) above seems to suggest four inflection points within the interval \([0,2\pi]\text{.}\)
Can we at least verify this algebraically? Yes, we can!
Let \(u=\sin x\text{,}\) so that \(f''(x)=-4u^2-u+2\text{.}\) We can solve \(f''(x)=0\) for \(u\) using the quadratic equation:
\begin{equation*}
u=\frac{-1\pm \sqrt{33}}{8}\text{.}
\end{equation*}
Thus we have a potential inflection point at points \(x\in [0,2\pi]\) satisfying
\begin{align*}
\sin x \amp =\frac{-1+\sqrt{33}}{8} \amp \sin x\amp =\frac{-1-\sqrt{33}}{8} \text{.}
\end{align*}
Each of the equations above has two solutions within \([0,2\pi]\text{,}\) and these correspond to the four inflection points we see in our graph of \(f\) over \([0,2\pi]\text{.}\)
