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#?

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Answer 1 · 2015-09-09T14:24:31

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

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#