Question - Analysis of a Trigonometric Function Graph
Solution:
\[
\begin{align*}
\text{Given } & f(x) = -4\sin x - \cos 2x, \text{ for } 0 \leq x \leq \pi.\\
\text{Find } f'(x) & = -4\cos x + \sin 2x \cdot 2 \text{ by using the chain rule.}\\
& = -4\cos x + 2\sin 2x \text{ where } \sin 2x = 2\sin x \cos x.\\
\text{Thus, } f'(x) & = -4\cos x + 4\sin x \cos x.\\
\text{For stationary points, set } f'(x) & = 0.\\
& -4\cos x + 4\sin x \cos x = 0.\\
& \cos x (-4 + 4\sin x) = 0.\\
\text{For } \cos x = 0, & \text{ we get } x = \frac{\pi}{2}.\\
\text{For } -4 + 4\sin x = 0, & \text{ we get } \sin x = 1.\\
& \text{No solution for } 0 \leq x \leq \pi \text{ as } \sin x = 1 \text{ only at } x = \frac{\pi}{2}.\\
\text{Stationary point at } & x = \frac{\pi}{2}.\\
\text{To classify this stationary point, find } f''(x).\\
& f''(x) = 4\sin x + 4\cos x \cdot \cos x - 4\sin^2 x.\\
\text{At } x = \frac{\pi}{2}, \text{ } & f''\left(\frac{\pi}{2}\right) = 4\cdot0 + 4\cdot0 - 4\cdot1 = -4.\\
\text{Since } f''\left(\frac{\pi}{2}\right) < 0, & \text{ the point is a maximum.}
\end{align*}
\]
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