Find the absolute maximum and minimum values of the function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur. I'll start typing the equation below in a reply.

Question

codysek98 · Answer

$$f(\theta) = \sin(\theta),  -(\frac{ 5\pi }{ 3 }) \le \theta \le-(\frac{ \pi }{ 2 })$$

solomonzelman · Answer

$f'(\theta)={\tiny~}?$

codysek98 · Answer

that's the derivative of f(theta), right?

solomonzelman · Answer

Yes.

codysek98 · Answer

f'(theta) = cos(theta)

is that correct?

solomonzelman · Answer

Yes, you are correct!

codysek98 · Answer

do i plug in -5pi/3 for the min and plug in -pi/2 for the max?

solomonzelman · Answer

So, the possible critical numbers, if the interval for $\theta$ is given are:

1.  Interval boundaries (only closed boundaries, not open boundaries).
2.  All values (within the given interval) where $f'(\theta)$ is undefined (provided $f(\theta)$ is defined at that value.

solomonzelman · Answer

So, yes, you plug in the interval boundaries.

codysek98 · Answer

cos(5pi/3) = .5

is that right?

solomonzelman · Answer

also,
3.   Any points where $f'(\theta)=0$.

solomonzelman · Answer

yes, you are right.

codysek98 · Answer

and then cos(-pi/2) = 0

solomonzelman · Answer

Yes

solomonzelman · Answer

And sin(-pi/2)=1, so you definitely have a local minimum at theta=-pi/2

solomonzelman · Answer

BUT, you plug in the boundaries into the function, not into the derivative.

codysek98 · Answer

oh okay. and how did you know for sure that there was a min at -pi/2 just because it equaled 1?

solomonzelman · Answer

Oh, I meant sin(-pi/2)=-1

solomonzelman · Answer

not positive 1, sorry.
And why did I know that? Because sine of any angle is always AT MOST +1, and AT LEAST -1.

solomonzelman · Answer

In other words,  $-1\le\sin[f(\theta)]\le1$.

codysek98 · Answer

oh okay. so the absolute minimum is -1 at -pi/2

solomonzelman · Answer

Yes, $f(\theta)=-1$, at $\theta=-\pi/2$.

codysek98 · Answer

i mean absolute minimum is -1 at theta = -pi/2

solomonzelman · Answer

yes, yes.

codysek98 · Answer

okay. so for the maximum, i now know the max would be positive one at theta = -5pi/3

solomonzelman · Answer

Actually, you are not correct.

codysek98 · Answer

oh wait is it .5?

solomonzelman · Answer

$\sin\theta$  has a period every $2\pi$.
So, if it has a minimum at $\theta=-\pi/2$, then it has other closest minimums at $\theta=3\pi/2$ and $\theta=-5\pi/2$.

solomonzelman · Answer

maximums of any sine function are right in between the minimums.

solomonzelman · Answer

https://www.desmos.com/calculator/kxfbopjxxp

codysek98 · Answer

so would the maximum be 1 at theta = -3pi/2 ?

solomonzelman · Answer

Yes.

codysek98 · Answer

thank you