`h(t) = 3t - arcsin(t)` Find the critical numbers of the function

Textbook Question

Chapter 4, 4.1 - Problem 42 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to find the critical points of the function, hence, you need to evaluate the solutions to the equation h'(t) = 0.

You need to evaluate the first derivative:

`h'(t) = 3 - 1/(sqrt(1-t^2))`

You need to solve for theta h'(t) = 0, such that:

`3 - 1/(sqrt(1-t^2))= 0 => 3(sqrt(1-t^2)) - 1 = 0 => (sqrt(1-t^2)) = 1/3`

Squaring both sides yields:

`1-t^2 = 1/9 => t^2 = 1 - 1/9 => t^2 = 8/9 => t_(1,2) = +-(2sqrt2)/3`

Hence, evaluating the critical numbers of the function for h'(t) = 0, yields  `t_(1,2) = +-(2sqrt2)/3.`

scisser | (Level 3) Honors

Posted on

`h(t) = 3t - arcsin( t ) `

Find the derivative

`h'(t) = 3 - ( 1 / sqrt( 1 - t² ) ) `

Set it equal to 0

`0 = 3 - ( 1 / sqrt( 1 - t² ) ) `

`1 / sqrt( 1 - t² ) = 3 `

`1 = 3sqrt( 1 - t² ) `

`1 / 3 = sqrt( 1 - t² ) `

`1 / 9 = 1 - t^2 `

`t^2 + ( 1 / 9 ) = 1 `

`t^2 = ( 8 / 9 ) `

`t = +-sqrt( 8 / 9 ) `

`t= +- (2sqrt(2))/3`