Consider the equation below. f(x) = 4x^3 + 6x^2 24x + 1 (a) Find the intervals o
ID: 2850116 • Letter: C
Question
Consider the equation below.
f(x) = 4x^3 + 6x^2 24x + 1
(a) Find the intervals on which f is increasing. (Enter your answer using interval notation.)
Find the interval on which f is decreasing. (Enter your answer using interval notation.)
(b) Find the local minimum and maximum values of f.
local minimum value
local maximum value
(c) Find the inflection point. (x, y) =
Find the interval on which f is concave up. (Enter your answer using interval notation.)
Find the interval on which f is concave down. (Enter your answer using interval notation.)
Explanation / Answer
f(x) = 4x^3 + 6x^2 24x + 1
f'(x) = 12x2 + 12x -24
for max or min, f'(x) = 0 => 12x2 + 12x -24 = 0 (x+2)(12x-12) = 0 => x = -2 or x = 1
f is increasing in (-infinity, -2) U (1 , infinity)
f is decresing in (-2, 1)
f''(x) = 24x + 12
f''(-2) = -36 <0, maximum occurs at x= -2
maximum value = f(-2) = 4(-2)^3 + 6(-2)^2 24(-2) + 1 = 41
f''(1) = 36 >0, minimum occurs at x=1
minimum value = 4x1^3 + 6x1^2 24 x 1 + 1 = -13
For point of infection: f''(x) = 0 => 24x + 12 = 0 => x = -0.5
concave down: (-infinity, -0.5)
concave up: (-0.5, infinity)