Local Max and Min and Inflection points?

Question

f(x) = 4x^3 + 18x^2 - 216x + 4

the first derivative test

f'(x)=12x^2+36x-216

zeros are -6 and 3

The graph increases from -inf to -6 and from 3 to inf

Decreases from -6 to -3

Wat are the local max and mins?

what are the inflection points

and where would the function be concave up or down

Answer 1

second derivative: f " = 24x + 36

f " = 0 at x = -1.5, which is the point of inflection

x < -1.5 ==> concave down (f " < 0)

x > -1.5 ==> concave up (f" > 0)

since you increase below -6 and decrease above it, there is a local max at x = -6 (find f(-6) to find the value of the local max))

you go from decreasing to increasing at x = 3, which puts a local min there.  find f(3) to find the value of the local min.

Answer 2

you plug in -6 and 3 back into f(x) to find which is greater for local max and mins.

take the second derviative and find the zeros for inflection points. you can find concavity by graphing or see if it's greater than zero for concave up and less than zero for regions for concave down.