Having a little bit of trouble figuring out this problem here.
Find the absolute maximum and absolute minimum values of $f(t)$ = $8t + 8 \cot(t/2)$, on the given interval [$π/4, 7π/4$].
So I know that I need to first take the derivative, which is $8+8\csc^2(t/2)*(1/2)$ and then let the $f'(t)$ $=$ $0$. I saw in a solution of another version that it just pertains to getting $(1/2)\csc^2(t/2)=0$ but I don't really understand why just that part, which then gets me lost in finding the absolute minimum/maximum.
Thanks.
Best Answer
You are correct that you need to set $f'(t)=0$. $$f'(t)=8-4\csc^2(t/2)=0$$ $$\csc(t/2)=\pm \sqrt2$$ $$t/2=\pi/4+2\pi k,-\pi/4+2\pi k$$ Where $k$ is any integer (such that $t$ is within the desired interval). So: $$t=\pi/2+4\pi k,-\pi/2+4\pi k$$ Then you just need to evaluate $f(t)$ at those points and the endpoints of your interval and find which ones are the maximum and minimum.
You can also take $f''(t)$ to see which ones might possibly be extremal points by seeing if they are a local maximum ($f''(t)<0$) or minimum ($f''(t)>0$): $$f''(t)=-4\cot(t/2)\csc^2(t/2)$$