f(x) = 8 + 2x^2- x4 (a) Find the interval of increase. (Enter your answer using

Question

f(x) = 8 + 2x^2- x4

(a) Find the interval of increase. (Enter your answer using interval notation.) Find the interval of decrease. (Enter your answer using interval notation.) (b) 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.) (c) Find the inflection points. (x, y) = (smaller x-value) (x, y) = (larger x-value) 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
f'(x) = 4x - 4x^3
f''(x) = 4 - 12x^2

(a)
for increasing, f'(x) > 0
4x - 4x^3 > 0
4x(1 - x^2) > 0
4x(x^2 - 1) < 0
4x(x - 1)(x + 1) < 0
x = (-inf, -1) U (0, 1)

for decreasing, f'(x) < 0
4x - 4x^3 < 0
4x(1 - x^2) < 0
4x(x^2 - 1) > 0
4x(x - 1)(x + 1) > 0
x = (-1, 0) U (1, inf)

(b)
for local minimum and local maximum, we find critical points
f'(x) = 0
4x - 4x^3 = 0
4x(1 - x^2) = 0
x = 0, 1, -1
f(0) = 8 + 2x^2 - x^4 = 8 + 0 - 0 = 8
f(1) = 8 + 2x^2 - x^4 = 8 + 2 - 1 = 9
f(-1) = 8 + 2x^2 - x^4 = 8 + 2 - 1 = 9
hence
local minimum value = 8
local maximum value = 9

(c)
for inflection poiint, f''(X) = 0
4 - 12x^2 = 0
3x^2 = 1
x = 1/sqrt(3), -1/sqrt(3)

for concave up, f''(x) > 0
4 - 12x^2 > 0
3x^2 - 1 < 0
x = (-1/sqrt(3), 1/sqrt(3))

for concave down, f''(x) < 0
4 - 12x^2 < 0
3x^2 - 1 > 0
x = (-inf, -1/sqrt(3) U (1/sqrt(3), inf)

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