Find (A) f'(x), (B) the partition numbers for f' , and (C) the critical numbers of f.

Question

f(x) = x³—12x+8

Answer 1

f(x) = x³ - 12x + 8

f'(x) = 3x² - 12  

 

I don't know what a partition number is, but I'm guessing you want to find where f' = 0 

 

3x² - 12 = 0

3(x² - 4) = 0

3(x-2)(x+2) = 0

 

So f' = 0 when x = 2  or  x = -2  

 

The critical numbers of f are x = 2 and x = -2

f(-2) = (-2)³ - 12(-2) + 8 = -8 + 24 + 8 = 24

f(2) = 2³ - 12(2) + 8 = 8 - 24 + 8 = -8 

 

The critical points are at (-2,24) and (2,-8) 

 

f''(x) = 6x, so by the 2nd derivative test (-2,24) is a local maximum and (2,-8) is a local minimum