Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 35

Given: f(x)=2+2x^2-x^4
Find the critical values by setting the first derivative equal to zero and solving for the x values.
f'(x)=4x-4x^3=0
4x(1-x^2)=0
x=0,x=-1,x=1
The critical values are x=0, x=-1, x=1.
Part a)
If f'(x)>0 the function is increasing in the interval.
If f'(x)<0 the function is decreasing in the interval.
Select an x value in the interval (- oo ,-1).
Since f'(-2)>0 the function is increasing in the interval (-oo ,-1).
Select an x value in the interval (-1, 0).
Since f'(-1/2)<0 the function is decreasing in the interval (-1, 0).
Select an x value in the interval (0, 1).
Since f'(1/2)> the function is increasing in the interval (0, 1).
Select an x value in the interval (1, oo ).
Since f'(2)<0 the function is decreasing in the interval (1, oo ).
Part b)
Because the function changes direction from increasing to decreasing a local maximum exist at x=-1 and x=1. The local maximum occurs at the points
(-1, 3) and (1, 3).
Because the function changes direction from decreasing to increasing a local minimum exists at x=0. The local minimum occurs at the point (0, 2).
Part c)
Find the critical values of the second derivative.
f''(x)=4-12x^2=0
4(1-3x^2)=0
1=3x^2
x=+-(sqrt(3)/3)=+-.577
If f''(x)>0 the function is concave up.
If f''(x)<0 the function is concave down.
If f"(x)=0 an inflection point exists.
Select an x value in the interval (-oo ,-.577).
Since f''(-1)<0 the function is concave down in the interval (-oo ,-.577).
Select an x value in the interval (-.577, .577).
Since f''(0)>0 the function is concave up in the interval ('.577, .577)
Select an x value in the interval (.577, oo ).
Since f''(1)<0 the function is concave down in the interval (.577, oo ).
Since f''( +- .577)=0 the function will have inflection points at x=+- .577.
The inflection points are at the coordinates (-.577, 2.555) and (.577,2.555).
Part d)