I have already solved both of these questions. Mainly, I just need somebody to check my work and tell me if it's correct. I'd really appreciate it!

**Problem 1:**
Find all local and global minimums and maximums of

$\displaystyle f(t) = 4t3 - 3t2 - 90t+21$

in the interval $\displaystyle -4 \leq t \leq 6$. List the coordinates of each below. If an item doesn't exist, write DNE.

**My Answer:**
First I found the critical points and classified them as local max's or min's.

$\displaystyle f\prime(x) = 12t^2 - 6t -90$

$\displaystyle 12t^2 - 6t-90 = 6(t-3)(2t + 5)$

$\displaystyle t = 3, -\frac{5}{2}$

Second Derivative Test

$\displaystyle f\prime\prime(x) = 24t - 6$

$\displaystyle f\prime\prime(3) = 24(3)-6 = 66 $

$\displaystyle f\prime\prime(-\frac{5}{2}) = 24(-\frac{5}{2}) - 6 = -66$

*So f(x) has a local max at *$\displaystyle (-\frac{5}{2}, 164.75)$

* and a local min at *$\displaystyle (3, -168)$

*.*
Then I tried to find the global extrema:

$\displaystyle f(-4) = 77$

$\displaystyle f(-\frac{5}{2}) = 164.75$

$\displaystyle f(3) = -168$

$\displaystyle f(6) = 237$

*So I said that f(x) has a global min of -168 and no global max on this interval *$\displaystyle ( -4 \leq x \leq 6 )$

Is all of that correct?

**Problem 2:**
Find the exact global maximum and minimum values of the function

$\displaystyle f(x) = sin(x) - xcos(x) - \frac{\pi}{2}cos(x)$

**My Answer:**
First I tried to find the critical points:

$\displaystyle f\prime(x) = sin(x)(x + \frac{\pi}{2})$

$\displaystyle sin(x)(x + \frac{\pi}{2}) = 0$

$\displaystyle x = -\frac{\pi}{2}, 0 , \pi$

At this point I realized that since there's no interval given, you could have infinitely many critical points with the unit circle, so I decided to graph the function. This is the graph:

After seeing this, I wrote the following as my answer:

*There is no global extrama because there is no interval and the graph keeps fluctuating bigger as x approaches negative and positive infinity.*
Is that correct? Is that a good way of wording it?

Thank you so much to anybody that replies!