`f(t) = t + cot(t/2), [(pi/4), ((7pi)/4)]` Find the absolute maximum and minimum values of f on the given interval

Textbook Question

Chapter 4, 4.1 - Problem 58 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

2 Answers | Add Yours

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

`f(t)=t+cot(t/2)`

differentiating,

`f'(t)=1-1/2csc^2(t/2)`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now to find the critical numbers, solve for t for f'(t)=0.

`1-1/2csc^2(t/2)=0`

`csc^2(t/2)=2`

`1/(sin^2(t/2))=2`

`sin^2(t/2)=1/2`

`sin(t/2)=+-1/sqrt(2)`

sin(t/2)=1/`sqrt(2)`   : t=`pi` /2 , 3`pi` /2 in the interval( `pi` /4 , 7`pi` /4)

sin(t/2)=-1/ `sqrt(2)`  : No solution

So the critical numbers are `pi` /2 and 3`pi` /2

Evaluating the function at the critical points and at the end points of the interval,

`f(pi/2)=pi/2+cotpi/4=pi/2+1=2.5708`

`f(3pi/2)=3pi/2+cot(3pi/4)=3pi/2-1=3.712`

`f(pi/4)=pi/4+cot(pi/8)=pi/4+1+sqrt(2)=3.19`

`f((7pi)/4)=(7pi)/4+cot(7pi/8)=(7pi)/4-1-sqrt(2)=3.083`

So the absolute maximum=(3`pi` /2-1) at t=3`pi` /2

and absolute minimum=(`pi` /2+1) at t=`pi` /2 

scisser's profile pic

scisser | (Level 3) Honors

Posted on

Find the derivative` `

`f'(t)=1-(1/2)csc^2(t/2) `

Set it equal to 0.

`0=1-(1/2)csc^2(t/2)`

`1=(1/2)csc^2(t/2)=1/(2sin^2(t/2)) `

`sin^2(t/2)=1/2 `

`sintheta=1/sqrt2 (for) pi/3, (3pi)/3, (9pi)/4 etc` ... `f'(t)=0` when

`t= pi/2, (3pi)/2`

Find the y-values of the critical values and endpoints

`f(pi/4)=3.2 `

`f(pi/2)= 2.6`

`f((3pi)/2)= 3.7`

`f((7pi)/4)= 3.1`

Therefore, the absolute min is at `t=pi/2 ` with a value of 2.6

The absolute max is at `t=(3pi)/2` with a value of 3.7

We’ve answered 318,917 questions. We can answer yours, too.

Ask a question