Prove that $g(x)=\frac{\sin(x)}{\sqrt x}$ has maximum and minimum on $(0,\pi]$

Question

Assume that $g(x)=\frac{\sin(x)}{\sqrt x}$ is given.

Prove that $g(x)$ has maximum and minimum on $(0,\pi]$ .

We've learned a theorem in class which i think is related :

Assume that $f:[a,b]\to\mathbb R$ is continuous. There exists $c,d \in [a,b]$ such that $\forall x \in [a,b] : f(c) \le f(x) \le f(d)$

Is is enough to prove that $g(x)$ is continous? ( $\sin(x)$ is continous on $(0,\pi]$ . $\sqrt x$ is not zero on that interval ... so, $g(x)$ is continuous too ... Is this correct? )

Answer 1

Observe that

$$f(x)=\begin{cases}f(x),&x\in(0,\pi]\\{}\\0,&x=0\end{cases}$$

is defined in $\;[0,\pi]\;$ and continuous there.

Answer 2

No, because for example the function $1/x$ is continuous on $(0,1]$ but it attains no maximum value in that interval.

However, you may consider the continuous extension of $g$ to the the bounded closed interval $[0,\pi]$. What is the limit $\lim_{x\to 0^+} g(x)$?

Moreover note that $g(x)\geq 0$ in $(0,\pi]$ and $g(\pi)=0$.