Second derivative test

Question

F(x)=x³-x⁴  second derivative test 

Answer 1

F(x) = x³ - x⁴

 

F'(x) = 3x² - 4x³ = x²(3-4x)

 

F'(x) = 0 when x = 0 or 3/4 ← critical points

 

F"(x) = 6x - 12x²

 

    F"(3/4) < 0   So, there is a relative max when x = 3/4

 

    F"(0) = 0  So, the Second Derivative Test is inconclusive.

 

       Using the First derivative Test, F'(x) > 0 when x < 0

 

                                                    F'(x) > 0 when 0 < x < 3/4

 

                 So, there is no relative extremum when x = 0.

 

Answer 2

F'(x) = 3x^2 - 4x^3

 

 F''(x) = 6x - 12x^2

 

 6x - 12x^2 = 0

 

6x( 1 - 2x) = 0

 

x=0 or x = 1/2

 

 

So for x<0, F''(x) <0 so concave down

for  0 < x < 1/2, F''(x)>0 so concave up

For x>1/2 , F''(x) <0 so concave down