Find the inflection point(s), if any, of the function.

Calculus
Tutor: None Selected Time limit: 1 Day

Jul 15th, 2015

Thank you for the opportunity to help you with your question!

First, we find the second derivative of the function.

g(x) = 4x^4 - 8x^3 + 2  -----------> g'(x) = (4x^4 - 8x^3 + 2)'  -----------> g'(x) = (4x^4)' - (8x^3)' + (2)'

g'(x) = 4(x^4)' - 8(x^3)' + (2)' ------------> g'(x) = 4(4x^4-1) - 8(3x^3-1) + 0 -------------> g'(x) = 4(4x^3) - 8(3x^2)

g'(x) = 16x^3 - 24x^2

g''(x) = (16x^3 - 24x^2)'  --------------> g''(x) = (16x^3)' - (24x^2)' --------> g''(x) = 16(3x^3-1) - 24(2x^2-1)

g''(x) = 48x^2 - 48x^1  ------------->  g''(x) = 48x^2 - 48x

Then we equal the second derivative to zero and solve for x.

48x^2 - 48x = 0  ------------> 48x(x - 1) = 0 . It is factored, so now we equal each factor to zero and solve for x like this:

48x = 0 ------> 48x/48 = 0/48 ---------> x = 0

x - 1 = 0  -------> x - 1 + 1 = 0 + 1  ---------> x = 1

Now we enter each x value into the original function g(x) in order to find the y coordinate like this.

g(0) = 4(0)^4 - 8(0)^3 + 2  ------------>   g(0) = 0 - 0 + 2 = 2. The ordered pair (x , y) is: (0 , 2)

g(1) = 4(1)^4 - 8(1)^3 + 2 -------------> g(1) = 4 - 8 + 2 = - 2. The ordered pair is: (1 ,  -2)

(0, 2)   smaller x

(1, -2) larger x

Please let me know if you have a doubt or question.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 15th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Jul 15th, 2015
...
Jul 15th, 2015
May 29th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer