`f(x) = x^3 - 6x^2 + 15` Find the critical numbers, open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

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Chapter 3, 3.3 - Problem 22 - Calculus of a Single Variable (10th Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Given: `f(x)=x^3-6x^2+15`

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

`f'(x)=3x^2-12x=0`

`3x(x-4)=0`

`x=0,x=4`

The critical value for the first derivative are x=0 and x=4.

If f'(x)>0, the function is increasing in the interval.

If f'(x)<0, the function is decreasing in the interval.

Choose a value for x that is less than 0.

f'(-1)=15 Since f'(-1)>0 the graph of the function is increasing in the interval (-oo,0).

Choose a value for x that is between 0 and 4.

f'(1)=-9 Since f'(1)<0 the graph of the function is decreasing in the interval

(0, 4).

Choose a value for x that is greater than 4.

f'(5)=15 Since f'(5)>0 the graph of the function is increasing in the interval

(4, `oo).`

Because the function changed direction from increasing to decreasing there will be a relative maximum at x=0. The relative maximum occurs at (0, 15).

Also, because the function changed direction from decreasing to increasing there will be a relative minimum at x=4. The relative minimum occurs at

(4, -17).

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