How do i find the extrema (local maximum and minimum) by using the first derivative of: x^3-(3/2)x^2?

Answer 1

Local minimum or maximum is points where first derivation=0

Derivation of x^3-(3/2)x^2 is

3x^2-(3/2)*2x=3x^2-3x
=3*(x^2-x)
=3*x*(x-1)

Zeros of this expresion is x=0 and x=1

Second derivate of this x^3-(3/2)x^2 is

6x-3

Function have local minimum where is second derivation is positive.

Function have local maximum where is second derivation is negative.

Test:
Second derivate=6x-3

x1=0 x2=1

For x=0 6x-3=6*0-3=0-3= -3 Local maximum

For x=1 6x-3=6*1-3=6-3=3 Local minimum

Answer 2

If you wat to see graph of your function in google type:
"function graphs online"

When you see list of results click on:

rechneronline.de/function-graphs

When page be open in blue rectacangle type:

x^3-1.5x^2

Then click option Draw

You will see graph of your function