Answer
$f(\theta)$ has the absolute maximum value of $1$ at $\theta=\pi/4$ and the absolute minimum value of $-\sqrt3$ at $\theta=-\pi/3$ in the defined domain.
Work Step by Step
$$f(\theta)=\tan\theta\hspace{1cm}-\frac{\pi}{3}\le\theta\le\frac{\pi}{4}$$
1) Find the derivative $f'(\theta)$: $$f'(\theta)=\sec^2\theta=\frac{1}{\cos^2\theta}$$
We do not have any values of $\theta$ for which $f'(\theta)=0$, but $f'(\theta)$ is undefined for $\cos\theta=0$, or $\theta=\pi/2+k\pi$ $(k\in Z)$. These are the critical points of $f(\theta)$.
However, there are no values of $\theta$ lying in the defined domain $[-\pi/3,\pi/4]$. So no critical points would be taken into consideration here.
2) Evaluate $f'(\theta)$ at the endpoints:
- For $\theta=-\pi/3$: $$f\Big(-\frac{\pi}{3}\Big)=\tan\Big(-\frac{\pi}{3}\Big)=-\sqrt3$$
- For $\theta=\pi/4$: $$f\Big(\frac{\pi}{4}\Big)=\tan\Big(\frac{\pi}{4}\Big)=1$$
This means $f(\theta)$ has the absolute maximum value of $1$ at $\theta=\pi/4$ and the absolute minimum value of $-\sqrt3$ at $\theta=-\pi/3$ in the defined domain.
