Let f'(x)=4x^3-12x^2 a. identify where the relative max extrema of f occur. b.Find the intervals on which f is decreasing and increasing. c.find where the graph of f is concave up concave down.

Answer 1

a)

First find the critical points

f'(x) = 0

4x³-12x² = 0

x²(x-3) = 0 , x = 0, 0, 3

Use the 2nd derivative test to determine relative max and min

f "(x) = 12x²-24x

f"(0) = 0 so that's a point of inflection

f"(3) = 36 > 0 so that's a relative minimum

b)

f(x) is decreasing from x =-∞ to 0 where it flattens and then continues to decrease to x=3

f(x) is increasing from x=3 to ∞

c) f(x) is concave up at x = 3

There is no concave down since x=0 is an inflection point