How do you find all relative extrema of the function f(x)= -x^3 -6x^2-9x-2?

Jul 27, 2015

Use the first derivative test and check for sign changes of ${f}^{'}$.

Explanation:

For a given function, relative extrema, or local maxima and minima, can be determined by using the first derivative test, which allows you to check for any sign changes of ${f}^{'}$ around the function's critical points.

For a critical point to be local extrema, the function must go from increasing, i.e. positive ${f}^{'}$, to decreasing, i.e. negative ${f}^{'}$, or vice versa, around that point.

So, start by determining the first derivative of $f$

${f}^{'} = - 3 {x}^{2} - 12 x - 9$

To determine the function's critical points, make ${f}^{'} = 0$ and solve for $x$

${f}^{'} = 0$

${f}^{'} = - 3 {x}^{2} - 12 x - 9 = 0$

This is equivalent to

$- 3 \left({x}^{2} + 4 x + 3\right) = 0$, or

${x}^{2} + 4 x + 3 = 0$

${x}_{1 , 2} = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot 3}}{2} = \left\{\begin{matrix}{x}_{1} = - 3 \\ {x}_{2} = - 1\end{matrix}\right.$

Since no domain restrictions are given for your function, both solutions will be critical points.

Now check to see if the first derivative changes sign around these points. Since you're dealing with two critical points, you're going to have to look at 3 intervals.

Select a value fro meach of these intervals and note the sign of ${f}^{'}$

• $\left(- \infty , - 3\right)$

${f}^{'} \left(- 4\right) = - 3 \cdot \left(- 4 + 1\right) \cdot \left(- 4 + 3\right)$

${f}^{'} \left(- 4\right) = - 3 \cdot \left(- 3\right) \cdot \left(- 1\right) = - 9 \to \textcolor{red}{\text{negative}}$

• $\left(- 3 , - 1\right)$

${f}^{'} \left(0\right) = - 3 \cdot \left(- 2 + 1\right) \cdot \left(- 2 + 3\right)$

${f}^{'} \left(0\right) = - 3 \cdot \left(- 1\right) \cdot \left(+ 1\right) = 3 \to \textcolor{g r e e n}{\text{positive}}$

• $\left(- 1 , \infty\right)$

${f}^{'} \left(0\right) = - 3 \cdot \left(0 + 1\right) \cdot \left(0 + 3\right)$

${f}^{'} \left(0\right) = - 3 \cdot 1 \cdot 3 = - 9 \to \textcolor{red}{\text{negative}}$

The first derivative changes sign twice. It goes from being negative to being positive around $x = - 3$, which means that this critical point is a local minimum.

On the other hand, it goes from being positive to being negative around point $x = - 1$, which means that this critical point is a local maximum.

This is equivalent to having a function that goes from decreasing to increasing (think of a valley) around point $x = - 3$, and from increasing to decreasing (think of a hill) around point $x = - 1$.

To get the actual points at which the function has the local minium and maximum, evaluate $f$ at the critical points.

$f \left(- 3\right) = - {\left(- 3\right)}^{3} - 6 {\left(- 3\right)}^{2} - 9 \left(- 3\right) - 2$

$f \left(3\right) = 27 - 54 - 27 - 2 = - 2$

and

$f \left(- 1\right) = - {\left(- 1\right)}^{3} - 6 {\left(- 1\right)}^{2} - 9 \left(- 1\right) - 2$
$f \left(- 1\right) = 1 - 6 + 9 - 2 = 2$

Therefore, the function $f$ has

$\textcolor{g r e e n}{\left(- 3 \text{,} - 2\right)} \to$ local minimum
$\textcolor{g r e e n}{\left(- 1 \text{,} 2\right)} \to$ local maximum

graph{-x^3 - 6x^2 - 9x - 2 [-10, 10, -5, 5]}