Question #cbfb8

1 Answer
Feb 27, 2017

There is one critical number, at #x= 1#, which is neither a maximum nor a minimum.

Explanation:

Start by finding the derivative by the power rule.

#f'(x) = 3x^2 - 6x + 3#

The critical points will occur when the derivative equals #0# or is undefined. This is a quadratic function, so it is defined on all of #x#.

#0 = 3x^2 - 6x + 3#

#0 = 3(x^2 - 2x + 1)#

#0 = (x -1)(x - 1)#

#x = 1#

Next, we must verify whether this is a local minimum or a local maximum. Note that this function will never have an absolute maximum/minimum, as shows the table below.

http://www.drcruzan.com/MathPolynomial.html

We can verify whether this point is a local min or a local max by finding the intervals of increase/decrease.

Test point 1: #x = 0#

#f'(0) = 3(0)^2 - 6(0) + 3 = 3#

Test point 2: #x = 3#

#f'(3) = 3(3)^2 - 6(3) + 3 = 27 - 18 + 3 = 12#

Since #f'(x) > 0# on both sides of #x = 1#, our single critical point is neither a maximum nor a minimum.

Hopefully this helps!