`g(x) = 2x^2 - 8x, [0,6]` Find the absolute extrema of the function on the closed interval.

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Chapter 3, 3.1 - Problem 19 - Calculus of a Single Variable (10th Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Given: `g(x)=2x^2-8x,[0,6]`

First find the critical x value(s) of the function. You will do this by setting the derivative function equal to zero and solving for the x value(s).

`g'(x)=4x-8=0`

`4x-8=0`

`x=2` 

Plug in the critical x value(s) and the endpoints of the closed interval into the g(x) function.

`g(x)=2x^2-8x`

`g(0)=2(0)^2-8(0)=0`

`g(2)=2(2)^2-8(2)=8-16=-8`

`g(6)=2(6)^2-8(6)=72-48=24`

Examine the g(x) values.

The absolute maximum value will be the point (6, 24).

The absolute minimum value will be the point  (2, -8).

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