`h(t) = t^(3/4) - 2t^(1/4)` Find the critical numbers of the function

Textbook Question

Chapter 4, 4.1 - Problem 37 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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mathace | (Level 3) Assistant Educator

Posted on

Given: `h(t)=t^(3/4)-2t^(1/4)`

Find the critical number(s) by setting the first derivative equal to zero and solving for the x value(s).

`h'(t)=(3/4)t^(-1/4)-(1/4)(2)t^(-3/4)=0`

`3/(4t^(1/4))-(1/(2t^(3/4)))=0` 

`3/(4t^(1/4))=(1/(2t^(3/4)))`

`6t^(3/4)=4t^(1/4)`

`6t^(3/4)-4t^(1/4)=0`

`2t^(1/4)[3t^(1/2)-2]=0`

`t=0`

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

`t^(1/2)=2/3`

`t=4/9`

The critical numbers are t=0 and t=4/9.

loves2learn's profile pic

loves2learn | (Level 3) Salutatorian

Posted on

take the derivative and set it equal to zero

`(3/4)t^(-1/4)-(1/2)t^(-3/4)=0 `


`3t^(-1/4)-2t^(-3/4)=0 `

`3t^(-1/4)=2t^(-3/4) `

raise both sides to the power -4

`3^(-4)t=2^(-4)t^3 `

`t/81=t^3/16 `
` t^3/16-t/81=0 `


`t(t^2/16-(1/81))=0 `

set each term equal to 0

`t=0 `

or

`t^2/16-(1/81)=0 `

so

`t=0 `

or

`t^2/16=1/81` 

`t^2=16/81`

`t=4/9 or -4/9 `


(but -4/9 doesn't work since in the original function you are taking a 4th root and you can do this with negative numbers)

So the answers are `t=0, 4/9 `

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