# How do you find the critical numbers for f(x) = 3x^4 + 4x^3 - 12x^2 + 5 to determine the maximum and minimum?

Nov 3, 2017

Critical points occur at $x \in \left\{- 2 , 0 , 1\right\}$

#### Explanation:

Critical point occur where the derivative of the function is equal to zero.

Given
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = 3 {x}^{4} + 4 {x}^{3} - 12 {x}^{2} + 5$

First derivative:
$\textcolor{w h i t e}{\text{XXX}} f ' \left(x\right) = 12 {x}^{3} + 12 {x}^{2} - 24 x$
which can be factored as
$\textcolor{w h i t e}{\text{XXX}} = \left(12\right) \left(x\right) \left({x}^{2} + 1 - 2\right)$

$\textcolor{w h i t e}{\text{XXX}} = \left(12\right) \left(x\right) \left(x + 2\right) \left(x - 1\right)$

Which implies the critical points (when $f ' \left(x\right) = 0$) occur when
$\textcolor{w h i t e}{\text{XXX}} x = 0$
$\textcolor{w h i t e}{\text{XXX")(x+2)=0color(white)("xxx}} \rightarrow x = - 2$ and
$\textcolor{w h i t e}{\text{XXX")(x-1)=0color(white)("xxx}} \rightarrow x = + 1$

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While not explicitly asked for in this question,
you can determine if each of these critical points is a minimum or maximum by evaluating the second derivative at each critical value.
Results greater than zero indicate a local minimum;
results less than zero indicate a local maximum;
results equal to zero indicate an inflection point.

{: (f''(x),=36x+24x-24,,), ("at "x=-2,= +72,>0,"minimum"),("at "x=0,=-24, <0, "maximum"),("at "x=+1,=+36,>0,"maximum") :}

Here is the graph for verification: 