`h(x) = x^5 - 5x + 2` Determine the open intervals on which the graph is concave upward or downward.

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Chapter 3, 3.4 - Problem 6 - 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^5-5x+2`

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

`f'(x)=5x^4-5`

`f''(x)=20x^3=0`

`x^3=0`

`x=0`

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

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

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

Choose a value for x that is less than 0.

f''(-1)=-20 Since f''(-1)<0 the graph is concave down in the interval (-`oo` ,0).

Choose a value of x that is greater than 0.

f''(1)=20 Since f''(1)>0 the graph is concave up in the interval (0, `oo).`

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