# What are the critical values, if any, of  f(x)=x^3 + 6x^2 − 15x?

Feb 21, 2016

I have found all the value for $x$ but left the finishing off for the question proposer

#### Explanation: $\textcolor{g r e e n}{\text{To find the x-axis intercepts}}$

Factorizing:" " x(x^2+6x-15)=0" "->" "color(blue)(x=0" is one solution")

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \text{ "->" } \frac{- 6 \pm \sqrt{36 - 4 \left(1\right) \left(- 15\right)}}{2 \left(1\right)}$

$x = \frac{- 6 \pm \sqrt{96}}{2} \text{ "=" } \frac{- 6 \pm 4 \sqrt{6}}{2}$

$x = - 3 \pm 2 \sqrt{6}$

$\textcolor{b l u e}{\implies x = - 7.899 \text{ to 3 decimal places}}$

$\textcolor{b l u e}{\implies x = + 1.899 \text{ to 3 decimal places}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{g r e e n}{\text{To find the Max/min}}$

$f ' \left(x\right) = 3 {x}^{2} + 12 x - 15 = 0$

$x = \frac{- 12 \pm \sqrt{144 - 4 \left(3\right) \left(- 15\right)}}{2 \left(3\right)}$

$x = - \frac{12}{6} \pm \sqrt{\frac{324}{36}}$

It is sometimes useful to take something inside the root as $\sqrt{\frac{324}{36}} = \sqrt{9}$

$x = - 2 \pm 3$

$\textcolor{b l u e}{\text{Max/Min at " x=-5 " and } x = 1}$

$\textcolor{g r e e n}{\text{I will let you finish this off}}$