Consider the equation below. (If you need to use -infinity or infinity, enter -I
ID: 3212200 • Letter: C
Question
Consider the equation below. (If you need to use -infinity or infinity, enter -INFINITY or INFINITY.) f(x) = 4x3 + 12x2 - 96x + 8 (a) Find the intervals on which f is increasing. (Enter the interval that contains smaller numbers first.) ( , ) union ( , ) Find the interval on which f is decreasing. ( , ) (b) Find the local minimum and maximum values of f. (max) (min) (c) Find the inflection point. ( , ) Find the interval on which f is concave up. ( , ) Find the interval on which f is concave down. ( , )Explanation / Answer
f(x) = 4x^3 + 12x^2 - 96x + 8
f'(x)=0
f'(x) = 12x^2 +24x -96 = 0
x^2 + 2x - 8 = 0
x^2 + 4x -2x -8 = 0
x(x+4) -2(x+4) =0
(x+4)(x-2) = 0
critical points are , x = -4 ,2
Maximum value f(-4) = 4(-4)^3 +12(-4)^2 -96(-4) +8 =328
Minimum value, f(2) = 4(2)^3 +12(2)^2 -96(2) +8 =-104