How do you find the interval where f is concave up and where f is concave down for f(x)= –(2x^3)–(3x^2)–7x+2?

1 Answer
Sep 9, 2015

Using the second derivative test, f(x) is concave up when x<-1/2 and concave down when x> -1/2.

Explanation:

Concavity has to do with the second derivative of a function.

A function is concave up for the intervals where d^2/dx^2f(x)>0.
A function is concave down for the intervals where d^2/dx^2f(x)<0.

First, let's solve for the second derivative of the function.

The first derivative:

f(x)=–2x^3–3x^2–7x+2

d/dx(–2x^3–3x^2–7x+2)

=-2(3)x^2-3(2)x-7

=-6x^2-6x-7

The second derivative:

f'(x) = -6x^2-6x-7

d/dx (-6x^2-6x-7)

=-6(2)x-6

=-12x-6

Intervals where f(x) is concave up:

-12x-6>0
-12x>6
x<-1/2

Intervals where f(x) is concave down:

-12x-6<0
-12x<6
x> -1/2