+0  
 
+5
143
1
avatar

what is the maximum and minimum for x^3-2x^2

Guest Feb 23, 2017
 #1
avatar+7093 
+1

y = x3 - 2x2

 

Take the derivative of this to find a function that will tell you the slope of the graph of that line at any point.

 

y' = 3x2 - 4x

 

Set y' = 0 to find all the points where the slope is 0.

 

0 = 3x2 - 4x

0 = (x)(3x - 4)

x = 0 and x = 4/3

 

So at x = 0 and at x = 4/3 the slope of x3 - 2x2 is 0. That means those points are maximums, minimums, or inflection points. To figure out which, take the second derivative.

 

y''= 6x - 4

 

Find y'' at x = 0

y'' = 6(0) - 4 = -4

Since y'' is negative at x=0, the graph is concave down at x = 0, so there is a max at x=0.

 

Find y'' at x = 4/3

y''= 6(4/3) - 4 = 8 - 4 = 4

Since y'' is positive at x = 4/3, the graph is concave up at x = 4/3, so there is a min at x = 4/3.

hectictar  Feb 23, 2017

6 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.