how many relative extrema and points of inflection does x^4-2x^3 have?

Question

What do you do when there is a common point where the both the first derivative and second derivative equal 0?

Answer 1

f = x^4 - 2x^3

f' = 4x^3 - 6x^2

4x^3 - 6x^2 = 0
2x^2 ( 2x - 3) = 0

2x^2 = 0
x = 0

2x-3 = 0
x = 3/2

f'' = 12x^2 - 12x

12x^2 - 12x = 0
12x ( x - 1 ) = 0

12x = 0
x = 0

x - 1 = 0
x = 1

Hence,

Relative minimum at ( 1/2 , -3/16 )
Inflection points = 0 , 1

When the critival value is 0 and the inflection value is 0, then, you use the inflection value to find the inflection point, not the critical value, though it won't matter, because you will be looking for the point where x = 0 by using the original function.

Answer 2

first derivative = 4x^3 - 6x^2 = 2x^2(2-3x)
=0 when x=0 or x = 2/3

2nd derivative = 12x^2 - 12x = 12x(x-1)
=0 when x= 0 or x=1

X=1 is an inflection point since the 2nd derivative changes sign when passing through x =1. It is negative for x<1 and +ve for x>1.

Here's a helpful link:

http://mathworld.wolfram.com/InflectionPoint.html