calculusderivatives
Hello everyone how would I find the extreme values of the following function.
$f(x)=\cos^2(x)$ within $0 \leq x \leq 2\pi$
I got the derivative as $f'(x)=-2\sin(x)\cos(x)=0$
I know that $\sin^{-1}(0)=0$ and $\cos^{-1}(0)=\pi/2$
But I am not sure if this is correct as the graph seems to go in a cycle so would there be critical points?
For any $x\in [0,2\pi]$ you get $f'(x)=\cos (x) -\sin (x)$ and $f''(x)=-\sin (x)-\cos (x)$. Therefore $$\begin{align} f'(x)=0&\iff \cos (x) -\sin(x)=0\\ &\iff \cos (x) =\sin (x)\\ &\iff x\in \{\frac{\pi}{4}, \frac{5\pi}{4}\} \end{align}$$
Which gives you the critical points $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.
The last equivalence is easy to see geometrically if you look at the unit circle.
I don't understand why you're messing with $\cos ^{-1}$. Care to explain so we can help you?
Similarly, for $f''$ you'll get the zeros $\displaystyle \frac{3\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$.
There are three possibilities for an extremum of a function on an interval:
A critical point is defined as case 2 or 3: you can't forget to check case 2 as well. In your case, $f'(x)$ is undefined both at $x=-1$ and $x=1$, so those are critical points, and they need to be checked to see if they are extrema.
Best Answer
The function $\cos^2 x$ is always $\ge 0$ and always $\le 1$. Find where equality holds and you have found the extreme values.