At which points does the graph of y = sqrt(sin x) have

At which points does the graph of $y = \sqrt{\sin x}$ have vertical tangents? Sketch the graph.

Step 1: Domain

The radicand must be non-negative, so the function exists only where $\sin x \geq 0$: $x \in \bigcup_{k\in\mathbb{Z}}\bigl[2k\pi,\ (2k+1)\pi\bigr]$.

The graph is a sequence of separate arches with gaps between them.

Step 2: Differentiate

By the chain rule, for $x$ strictly inside an arch: $y' = \frac{\cos x}{2\sqrt{\sin x}}$.

Step 3: Locate the vertical tangents

A vertical tangent occurs where $|y'| \to \infty$. The numerator $\cos x$ is $\pm 1$ at the ends of each arch while the denominator tends to zero, so both endpoints qualify:

\[ \lim_{x\to(2k\pi)^{+}}\frac{\cos x}{2\sqrt{\sin x}} = \frac{1}{0^{+}} = +\infty, \qquad \lim_{x\to((2k+1)\pi)^{-}}\frac{\cos x}{2\sqrt{\sin x}} = \frac{-1}{0^{+}} = -\infty. \]

So the graph meets the $x$-axis vertically at every multiple of $\pi$.

Step 4: The graph

Each arch rises from $0$ with infinite slope, peaks at $y = 1$ when $x = \tfrac{\pi}{2}+2k\pi$, and falls back to $0$ with infinite slope; the dashed lines mark the vertical tangents.

Step 5: Verification

\[x = \frac{\pi}{2}:\quad y = 1,\quad y' = \frac{\cos\frac{\pi}{2}}{2\sqrt{1}} = 0, \text{ the horizontal tangent at the peak} \;\;\checkmark\]

\[x = 0.01:\quad y' = \frac{\cos 0.01}{2\sqrt{\sin 0.01}} \approx \frac{0.99995}{2(0.099998)} \approx 5.0 \text{, already large} \;\;\checkmark\]