`g(theta) = 4theta - tan(theta)` Find the critical numbers of the function

Textbook Question

Chapter 4, 4.1 - Problem 40 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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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 `g'(theta) = 0` .

You need to evaluate the first derivative:

`g'(theta) = 4 - 1/(cos^2 theta)`

You need to solve for `theta` `g'(theta) = 0,` such that:

`4 - 1/(cos^2 theta) = 0 => 4(cos^2 theta) - 1 = 0 => 4(cos^2 theta) = 1`

`(cos^2 theta) =1/4 => cos theta = +-1/2`

`cos theta = 1/2 => theta = +-arccos(1/2) + 2kpi`

`theta = +-pi/3 + 2kpi`

`cos theta = -1/2 => theta = pi+-arccos(1/2) + 2kpi`

`theta = pi + pi/3 + 2kpi => theta = (4pi)/3 + 2kpi`

or

`theta = pi - pi/3 + 2kpi => theta = (2pi)/3 + 2kpi`

Hence, evaluating the critical numbers of the function for `g'(theta) = ` 0, yields `theta = +-pi/3 + 2kpi, theta = (4pi)/3 + 2kpi, theta = (2pi)/3 + 2kpi.`

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scisser | (Level 3) Honors

Posted on

`g(theta) = 4theta - tan(theta) `

`g'(theta) = 4 - sec^2(theta) `

equating f ' (x) to zero

`4 - sec^2(theta) = 0 `

`sec^2(theta) = 4 `

`sec theta = 2 and -2 `

`cos theta = 1/2 and - 1/2 `

` theta = pi/3, (2pi)/3, (4pi)/3, (5pi)/3, (7pi)/3, (8pi)/3, (10pi)/3 and (11pi)/3`

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