# `f(x) = (1/2)x^4 + 2x^3` Find the points of inflection and discuss the concavity of the graph of the function.

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### 1 Answer

Given: `f(x)=(1/2)x^4+2x^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).

`f'(x)=2x^3+6x^2`

`f''(x)=6x^2+12x=0`

`6x(x+2)=0`

`x=0,x=-2`

The critical value for the second derivative is x=0 and x=-2.

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 -2.

f''(-3)=18 Since f''(-3)>0 the graph is concave up in the interval (-`oo,-2` ).

Choose a value for x that is between -2 and 0.

f''(-1)=-6 Since f''(-1)<0 the graph is concave down in the interval (-2, 0).

Choose a value for x that is greater than 0.

f''(1)=18 Since f''(1)>0 the graph is concave up in the interval (0, `oo).`

Because the direction of concavity changes twice and because f''(-2)=0 and

f"(0)=0 there will be an inflection point at x=-2 and x=0.

The inflection points are (-2, -8) and (0, 0).