`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).
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gsarora17 | (Level 2) Associate Educator

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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.






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,





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

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

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scisser | (Level 3) Honors

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Find the derivative` `

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

Set it equal to 0.


`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

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