f(x) = x4-50x2 + 2 (a) Find the interval on which f is increasing. (Enter your a
ID: 2866194 • Letter: F
Question
f(x) = x4-50x2 + 2
(a) Find the interval on which f is increasing. (Enter your answer in interval notation.) Find the interval on which f is decreasing. (Enter your answer in interval notation.) (b) Find the local minimum and maximum values of f. local minimum local maximum (c) Find the inflection points. (x, y) = (smaller x-value) (x, y) = (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) = x^4 - 50x^2 + 2
f'(x) = 4x^3 - 100x
f''(x) = 12x^2 - 100
(a)
for increasing, f'(x) > 0
4x^3 - 100x > 0
4x(x^2 - 25) > 0
4x(x - 5)(x + 5) > 0
x = (-5, 0) U (5, inf)
for decreasing, f'(x) < 0
4x^3 - 100x < 0
4x(x^2 - 25) < 0
4x(x - 5)(x + 5) < 0
x = (-inf, -5) U (0, 5)
(b)
to find local minimum and maximum value, first we find critical points
put f'(x) = 0
4x^3 - 100x = 0
4x(x^2 - 25) = 0
4x(x - 5)(x + 5) = 0
x = 0, -5, 5
f(0) = x^4 - 50x^2 + 2 = 2
f(-5) = x^4 - 50x^2 + 2 = 625 - 1250 + 2 = -623
f(5) = x^4 - 50x^2 + 2 = 625 - 1250 + 2 = -623
hence
local minimum value = -623
local maximum value = 2
(c)
for inflection points, f''(x) = 0
12x^2 - 100 = 0
x^2 = 100/12 = 25/3
x = +5 / sqrt(3), -5 / sqrt(3)
for concave up, f''(x) > 0
12x^2 - 100 > 0
x = (-inf, -5 / sqrt(3)) U (5 / sqrt(3), inf)
for concave down, f''(x) < 0
x = (-5 / sqrt(3), 5 / sqrt(3))
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