`f(x) = -x^3 + 6x^2 - 5` Find the points of inflection and discuss the concavity of the graph of the function.

Textbook Question

Chapter 3, 3.4 - Problem 16 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Given: `f(x)=-x^3+6x^2-5`

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)=-3x^2+12x`

`f''(x)=-6x+12=0`

`-6x=-12`

`x=2`

The critical value for the second derivative is 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''(0)=12 Since f''(0)>0 the graph is concave up in the interval (`-oo` 2).

Choose a value for x that is greater than 2.

f''(3)=-6 Since f''(3)<0 the graph is concave down in the interval (2, `oo).`

Since the concavity changed directions and f''(2)=0 the inflection point occurs at x=2. The inflection point is (2, 11).