Consider the equation below. f(x) = x^4 -8x^2 + 8 (a) Find the interval on which
ID: 2837935 • Letter: C
Question
Consider the equation below.
f(x) = x^4 -8x^2 + 8 (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) (- 2,0),(2,infinity) Find the interval on which f is decreasing. (Enter your answer in interval notation.) ( -infinity , -2),(0,2) (b) Find the local minimum and maximum values of f. local minimum -8 local maximum 8 (c) Find the inflection points. (x, y) = ( X ) (smaller x-value) (x, y) = ( X ) (larger x-value) Find the interval on which f is concave up. (Enter your answer in interval notation.) Find the interval on which f is concave down. (Enter your answer in interval notation.)Explanation / Answer
f(x) = x4 - 8x2 + 8
f'(x) = 4x3 - 16x
f''(x) = 12x2 - 16
(a)
for increasing f'(x) > 0
4x3 - 16x > 0
4x(x2 - 4) > 0
4x(x - 2)(x + 2) > 0
x = (-infinity to -2) U (0, 2)..............for increasing
x = (-2, 0) U (2, infinity)...........for decreasing
(b)
for critical points, f'(x) = 0
4x(x - 2)(x + 2) = 0
x = -2, 0, 2
f''(x) = 12x2 - 16
f''(-2) = 48 - 16 = 32 > 0.........local minimum
f''(0) = -16 < 0.............local maximum
f''(2) = 48 - 16 = 32 > 0.......local minimum
f(-2) = 16 - 64 + 8 = -40
f(0) = 0 - 0 + 8 = 8
f(2) = 16 - 64 + 8 = -40
Hence local minimum value is (-40)
local macimum value is 8
(c)
for inflection points, f''(x) = 0
12x2 - 16 = 0
x2 = 4/3
x = 2/sqrt(3)........larger x-value
x = -2/sqrt(3).....smaller x-value
for concave up, f''() > 0
12x2 - 16 > 0
12(x2 - 4/3) > 0
x = (-infinity, -2/sqrt(3)) U (2/sqrt(3), infinity).......concave up
x = (-2/sqrt(3), 2/sqrt(3))......concave down