f(x)=x^4-8x^2+5
a] To get the intervals required we proceed as follows:
find the derivative of the f
f'(x)=4x^3-16x
set the derivative equal to 0
4x^3-16x=0
4x^3=16x
4x^3=16x
x^2=4
x=+/-2
x=-2 or x=2
These are our critical numbers.
Pick a value that falls in the region -2, 2 and plug it into the first derivative, and not whether the result is positive or negative:
for example use -3,-1,0,1 and 3
f'(x)=4x^3-16x
f'(-3)=-60
f'(-1)=12
f'(0)=0
f'(1)=-10
f'(3)=60
These results are respectively, negative, positive, negative, postive
This tells us that the graph is increasing where the derivative is positive and vice versa.
This tells us that the function switches from decreasing to increasing at -2, then the function switches from decreasing to increasing at 2.
The sign switches from positive to negative at 0
The graph increase at the interval [-2,0]∪[2,∞]
The graph decreases at the interval (-∞,-2]∪[0,2]
b]
The local maximum is at x=0
The local minimum is at x=-2 and x=2
c] The graph has no point of inflation.
