Question

find the critical numbers of the function. (enter your answers as a comma - separated list.)
$h(x)=\sin ^{2}x+\cos x\quad 0 < x < 2\pi$
$x=$

Explanation:

Step1: Differentiate the function

Using the chain - rule, if \(y = \sin^{2}x=( \sin x)^{2}\), then \(y^\prime = 2\sin x\cos x\) (by \(u = \sin x\), \(y = u^{2}\), \(y^\prime=2u\cdot u^\prime\)) and the derivative of \(\cos x\) is \(-\sin x\).
So \(h^\prime(x)=2\sin x\cos x-\sin x=\sin x(2\cos x - 1)\)

Step2: Set the derivative equal to zero

Set \(h^\prime(x)=0\), then \(\sin x(2\cos x - 1)=0\)

• Case 1: \(\sin x=0\)

Since \(0\lt x\lt2\pi\), then \(x = \pi\) (because \(\sin x = 0\) when \(x = k\pi,k\in\mathbb{Z}\), and in the interval \((0,2\pi)\), \(k = 1\) gives \(x=\pi\))

• Case 2: \(2\cos x-1 = 0\)

Solve \(2\cos x-1 = 0\) for \(x\). First, rewrite it as \(\cos x=\frac{1}{2}\)
Since \(\cos x=\frac{1}{2}\) and \(0\lt x\lt2\pi\), then \(x=\frac{\pi}{3}\) or \(x = 2\pi-\frac{\pi}{3}=\frac{5\pi}{3}\) (because \(\cos x\) is positive in the first and fourth quadrants and \(\cos\alpha=\cos(2\pi - \alpha)\))

Answer:

\(\frac{\pi}{3},\pi,\frac{5\pi}{3}\)