Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 38

Given: h(x)=5x^3-3x^5
Find the critical numbers by setting the first derivative equal to zero and solving for the x values.
h'(x)=15x^2-15x^4=0
15x^2(1-x^2)=0
x=0,x=1,x=-1
The critical values are x=0, x=1, x=-1.
Part a)
If h'(x)>0 the function increases in the interval.
If h'(x)<0 the function decreases in the interval.
Select an x value in the interval (-oo ,-1).
Since h'(-2)<0 the function is decreasing in the interval (-oo ,-1).
Select an x value in the interval (-1, 0).
Since h'(-1/2)>0 the function increases in the interval (-1, 0).
Select an x value in the interval (0, 1).
Since h'(1/2)>0 the function increases in the interval (0, 1).
Select an x value in the interval (1, oo ).
Since h'(2)<0 the function decreases in the interval (1,oo ).
Because the function changes direction from decreasing to increasing a local minimum exists at x=-1. The local minimum occurs at the point (-1, -2).
Because the function changes direction from increasing to decreasing a local maximum exists at x=1. The local maximum occurs at the point (1, 2).
Part c
h''(x)=30x-60x^3=0
30x(1-2x^2)=0
x=0, x=+-(sqrt(2))/(2)=+-.707
The critical values of the second derivative are x=0, x=-.707, and x=.707
If h''(x)>0 the graph of the function is concave up.
If h''(x)>0 the graph of the function is concave down.
If h''(x)=0 an inflection point exists.
Select an x value in the interval (-oo , -.707).
Since h''(-1)>0 the function is concave up in the interval (-oo ,-.707).
Select an x value in the interval (-.707, .707).
Since h''(0)=0 an inflection point will occur at x=0. The inflection point occurs at the coordinate (0, 0).
Select an x value in the interval (.707, oo ).
Since h''(1)<0 the function is concave down in the interval (.707, oo ).
Part d)