# Absolute min and max

• March 18th 2013, 05:37 PM
Oldspice1212
Absolute min and max
Find the absolute minimum and absolute maximum values of f on the given interval.

f(t) = 3t + 3 cot(t/2),
[π/4, 7π/4]

so
the derivative is -3/2cos(t)csc^2(t/2)

Then I make it = 0, etc,etc.

• March 18th 2013, 06:48 PM
Soroban
Re: Absolute min and max
Hello, Oldspice1212!

Your derivative is wrong . . .

Quote:

$\text{Find the absolute minimum and absolute maximum values of }f\text{ on the given interval.}$

. . $f(t) \:=\:3t + 3\cot(\tfrac{t}{2}),\;\;[\tfrac{\pi}{4},\:\tfrac{7\pi}{4}]$

We have: . $f'(x) \;=\;3 - 3\csc^2(\tfrac{t}{2})\!\cdot\!\tfrac{1}{2}$

. . . $3 - \tfrac{3}{2}\csc^2(\tfrac{t}{2}) \;=\;0$

. . . . $2 - \csc^2(\tfrac{t}{2}) \;=\;0$

n . . . . . $\csc^2(\tfrac{t}{2}) \;=\;2$

. . . . . . . $\csc(\tfrac{t}{2}) \;=\;\pm\sqrt{2}$

. . . . . . . . . . $\frac{t}{2} \;=\;\pm\frac{\pi}{4},\:\pm\frac{3\pi}{4},\:\pm \frac{5\pi}{4}\:\hdots$

. . . . . . . . . . $t \;=\;\pm\frac{\pi}{2},\:\pm\frac{3\pi}{2},\:\pm \frac{5\pi}{2}\:\hdots$

Therefore: . $t \;=\;\left\{\frac{\pi}{2},\:\frac{3\pi}{2}\right\}$

I'll let you calculate the maximum and minimum.
• March 18th 2013, 08:18 PM
Oldspice1212
Re: Absolute min and max
Ok so for pi/2 I got 3+3pi/2, for pi/4 I got 3pi/4+3cot(pi/8), 3pi/2 I got 9pi/2-3, and for 7pi/4 I got 21pi/4-3cot(pi/8)? So is the 3+3pi/2 min and 9pi/2-3 the max? Thanks a lot btw.
• March 19th 2013, 07:40 AM
HallsofIvy
Re: Absolute min and max
Note that the absolute min and max on a closed bounded interval may be at the endpoints of the interval even if the derivative is not zero there.