`y = 3cos(x), [0,2pi]` Find the absolute extrema of the function on the closed interval.

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Chapter 3, 3.1 - Problem 35 - Calculus of a Single Variable (10th Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Given: `y=3cos(x),[0,2pi]`

First find the critical x values of the function. To find the critical values, set the derivative equal to zero and solve for the x value(s). 

`f'(x)=-3sin(x)=0`

`-3sin(x)=0`

`sin(x)=0` 

`x=0,pi,2pi`

Plug in the critical values and the endpoints of the closed interval in to the original function.

`f(x)=3cos(x)`  

`f(0)=3cos(0)=3(1)=3`

`f(pi)=3cos(pi)=3(-1)=-3`

`f(2pi)=3cos(2pi)=3(1)=3`

Examine the f(x) values.

The absolute maximum is at the points  `(0,3) (2pi,3)`

The absolute minimum is at the point `(pi,-3)`

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