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

Guest Feb 23, 2017
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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

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