Thread: max min value of function [pi/4, 7pi/4]

1. max min value of function [pi/4, 7pi/4]

Find absolute max and absolute min of the function on the interval $\displaystyle [\frac{pi}{4}, \frac{7pi}{4}]$

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

I found the derivative:

$\displaystyle f'(t) = 1 -csc^2(t)*\frac{1}{2}$

Now need to find the critical numbers:

$\displaystyle f'(t) = 1 -csc^2(t)*\frac{1}{2} = 0$

2. Re: max min value of function [pi/4, 7pi/4]

Originally Posted by rabert1
Find absolute max and absolute min of the function on the interval $\displaystyle [\frac{pi}{4}, \frac{7pi}{4}]$

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

I found the derivative:

$\displaystyle f'(t) = 1 -csc^2(t)*\frac{1}{2}$

Now need to find the critical numbers:

$\displaystyle f'(t) = 1 -csc^2(t)*\frac{1}{2} = 0$
$\displaystyle \csc^2\left(\frac{t}{2}\right) = 2$

$\displaystyle \csc\left(\frac{t}{2}\right) = \pm \sqrt{2}$

$\displaystyle \sin\left(\frac{t}{2}\right) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

unit circle values, right? proceed ...

... edited for $\displaystyle \frac{t}{2}$ as the trig function argument.

3. Re: max min value of function [pi/4, 7pi/4]

Just wanted to add that $\displaystyle f'(t) = 1-\frac{\csc^2(t/2)}{2}$.