Find extrema on the interval

Question

Problem

Find the extrema of the function $$f(x) = cos^2(x)$$ on the interval $ [-4,4]$

I can differentiate and get $$f'(x) = -2 \sin(x) \cos(x)$$

And set that to zero, but I'm pretty sure that's not entirely correct.

Answer 1

You've done correct, the exterma are found where the derivative is zero. Observe that: $$f'(x) = -2\sin(x)\cos(x) = -\sin(2x)$$

And find the solutions for $\sin(2x) = 0$. That is, $2x = n\pi$ for any integer $n$ so that the value is inside the interval $[-4,4]$.

Answer 2

Hint: What you've done is right - setting $f^\prime=0$. Find the solutions, say $x,y$. Then, choose a small $\epsilon$. Calculate $$f^\prime(x+\epsilon),\quad f^\prime(x-\epsilon),\quad f^\prime(y+\epsilon)\quad f^\prime(y-\epsilon)$$ See if there is a sign change. Also, use the fact that $2\sin x\cos x=\sin(2x)$.