Finding the Absolute Max & Min Values

Question

This problem is giving a bit of trouble. This is for my Calculus one class and am trying to better understand this problem. Any help would be much appreciated.

$f(t) = 6t + 6 \cot (\frac{t}{2})$

The given intervals are:

$ [ \frac {\pi}{4},\frac{7\pi}{4}]$

Here's what I have:

I found that $f'(t) = 6 - \frac{1}{2}\csc^2 (\frac{t}{2}) = 0$

then I got to this point and drew a blank:

$\csc(\frac {t}{2}) = \sqrt{12}$

Answer 1

There's an arror in $f'(t)$. It should be $6(1-\frac{1}{2}\csc^2\frac{t}{2})$. So,

$$\csc\frac{t}{2} = \pm\sqrt{2}$$

Now, $\frac{t}{2}\in(\frac{\pi}{8},\frac{7\pi}{8}), \csc(\frac{t}{2})>0$

$$\sin\frac{t}{2} = \frac{1}{\sqrt 2}$$

So we have $\frac{t}{2} = \frac{\pi}{4}$ or $\frac{3\pi}{4}$.

Now check for extrema.

Answer 2

Don't forget to use the chain rule when taking the derivative of $cot(\frac{t}{2})$. You have to multiply $-csc^2(\frac{t}{2})$ by the derivative of $\frac{t}{2}$, which you can find using quotient rule or constant multiple rule since $\frac{t}{2}=\frac{1}{2}t$. This should help get you back on the right track!