Question

List the relative extrema and inflection points for f(x) = x3/3 +6x2 +1

Answer 1

what do you mean by extrema?

Regarding inflection points you take the second derivative and equate it to zero, I think this would give you the inflection points. (You should only get one as this is a cubic function).

Answer 2

In this case, "extrema" signifies minima and maxima, relative or local (not absolute).

To find these extrema, take the first derivative of the given function\[f(x)=x^3/3 +6x^2 +1\] set this derivative = to 0, and solve for the x-values for which the derivative is zero. These x-values are called "critical values."

Seeing that you're learning how to find and use the 2nd derivative: Once you have the critical values, substitute each one (separately) into the 2nd derivative.
Rules:
If the 2nd deriv. is + at a critical value, the curve is concave UP there, and thus you have a relative minimum.
If, however, the 2nd deriv. is (-) at such a vallue, the curve is conc. DOWN there, and thus you have a real. max.

Try finishing this problem. Optionally, graph the curve, showing the critical values, the direction of concavity for each c. v., and labeling the relative or local extrema.