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Question #cbfb8

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Answer 1 · 2017-02-27T03:16:41

There is one critical number, at $x = 1$ $x= 1$ , 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!

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