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#?

Question

Isabella Ackerman · Accepted Answer

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#

Scarlett Allison · Answer

undefined To find the intervals where $ f(x) = -(2x^3) - (3x^2) - 7x + 2 $ is concave up and concave down, we first need to determine its second derivative, $ f''(x) $. Then, we identify where $ f''(x) $ is positive (indicating concave up) and where it is negative (indicating concave down). First, let's find $ f''(x) $: $$ f'(x) = -6x^2 - 6x - 7 $$ $$ f''(x) = -12x - 6 $$ Now, to find where $ f''(x) $ changes sign, we set it equal to zero and solve for $ x $: $$ -12x - 6 = 0 $$ $$ x = -\frac{6}{12} = -\frac{1}{2} $$ Now, we test the intervals $ x < -\frac{1}{2} $, $ x = -\frac{1}{2} $, and $ x > -\frac{1}{2} $ by picking test points. For $ x < -\frac{1}{2} $, let's pick $ x = -1 $: $$ f''(-1) = -12(-1) - 6 = 6 $$ (Positive, so concave up) For $ x > -\frac{1}{2} $, let's pick $ x = 0 $: $$ f''(0) = -12(0) - 6 = -6 $$ (Negative, so concave down) Thus, $ f(x) $ is concave up for $ x < -\frac{1}{2} $ and concave down for $ x > -\frac{1}{2} $.