Consider the function below. f(x) = 8 + 2x^2 - x^4 Find the interval of increase
ID: 2873759 • Letter: C
Question
Consider the function below. f(x) = 8 + 2x^2 - x^4 Find the interval of increase. (Enter your answer using interval notation.) Find the interval of decrease. (Enter your answer using interval notation.) Find the local minimum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Find the local maximum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Find the inflection points. Find the interval where the graph is concave upward. (Enter your answer using interval notation.) Find the interval where the graph is concave downward. (Enter your answer using interval notation.)Explanation / Answer
f(x) = 8+ 2x^2 - x^4
a.
f'(x) = 4x - 4x^3
4x - 4x^3 = 0
x - x^3 = 0
x(1 - x^2) = 0
=> x = 0
=> 1- x^2 = 0
x^2 = 1
x = ±1
f'(-2) = -8 + 32 > 0
f'(2) = 8- 32 < 0
So f is increasing on (-1,0)and decreasing on (0,1)
b.
Following from (a), we have a minimum at f(0) = (0,8).
Maxima occur at (-1,f(-1)) = (-1,9) and (1,f(1)) = (1,9).
c.
f''(x) = 4 - 12x^2
4 - 12x^2 = 0
1 - 3x^2 = 0
3x^2 = 1
x^2 = 1/3
x = ±(1/3)
f''(-1) = 1 - 3 < 0
f''(0) = 1 > 0
f''(1) = 1 - 3< 0
So inflection points occur at (-(1/3),f(-(1/3))) = (-(1/3),77/4) and ((1/3),f((1/3))) = ((1/3),77/4).
The graph is concave-down on (-,-(1/3))((1/3),).
It is concave-up on (-(1/3),(1/3)).