Having difficulty with this one

Question

Find all critical numbers for the function f(x)=x^3-12x^2

Answer 1

f'(x) = 3x²- 24x = 3x(x-8) = 0 has zeros 0 & 8; these are the critical numbers.

 

The numberline can help you analyze f' in the three intervals:

 

x:  .......0.............8......................

f':    +   0      -      0  +

f:   incr       decr        incr                (increasing/decreasing over the above intervals)

 

This chart enables you to include that there is a local maximum at x=0 and a local minimum at x=8.

Answer 2

A point x=c is a critical point of f(x) if f(c) exists and f'(c) = 0 or does not exist.  So take the derivative of f(x) wrt x, set it to zero, and solve for x.  The solutions are the critical points if f(x) exists at those points.

 

f(x) = x³ - 12x²

 

df(x)/dx = 3x² - 24x

0 = 3x² - 24x

0 = 3x(x-8)

x = 0 and 8 are the critical points.  f(0) = 0 and f(8) = -256 so f(x) exists at both points.