# `h(x) = 12x - x^3` Determine the open intervals on whcih the graph is concave upward or downward.

### Textbook Question

Chapter 3, 3.4 - Problem 4 - Calculus of a Single Variable (10th Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Given: `h(x)=12x-x^3`

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

`h'(x)=12-3x^2`

`h''(x)=-6x=0`

`x=0`

The critical value for the second derivative is x=0.

If h''(x)>0, the curve is concave up in the interval.

If h''(x)<0, the curve is concave down in the interval.

Choose a value for x that is less than 0.

h''(-1)=6 Since h''(-1)>0 the graph is concave up in the interval (-oo,0).

Choose a value for x that is greater than 0.

h''(1)=-6 Since h''(1)<0 the graph is concave down in the interval (0, `oo).`

Because the function changed from concave up to concave down and h''(0)=0, there will be and inflection point and x=0

The inflection point is (0, 0).