#1**+1 **

y = x^{3} - 2x^{2}

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' = 3x^{2} - 4x

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

0 = 3x^{2} - 4x

0 = (x)(3x - 4)

x = 0 and x = 4/3

So at x = 0 and at x = 4/3 the slope of x^{3} - 2x^{2} 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