How do use the first derivative test to determine the local extrema F(x) = -2x^3 - 9x^2 + 24x + 40?

Jul 27, 2015

Determine the critical points of the function and check to see if the first derivative changes sign around these points.

Explanation:

The first derivative test allows you to determine whether or not a critical point of a function is also a local minimum or a local maximum.

This can be done by checking to see if the first derivative of the function changes signs around these critical points.

The idea is that a critical point is a local minimum if the function goes from decreasing to increasing, i.e. if ${f}^{'}$ goes from negative to positive*, around that point.

Likewise, a critical point is a local maximum if the function goes from increasing to decreasing, i.e. if ${f}^{'}$ goes from positive to negative, around that point.

So, start by determining the first derivative for your function

${f}^{'} = - 6 {x}^{2} - 18 x + 24$

This is equivalent to

${f}^{'} = - 6 \left({x}^{2} + 3 x - 4\right)$

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

${f}^{'} = - 6 \left({x}^{2} + 3 x - 4\right) = 0$

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

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

Since you have no domain restrictions for this function, both solutions will be critical points.

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

Select a value from each of these intervals and use it to determine the sign of ${f}^{'}$ for those respective intervals.

• $\left(- \infty , - 4\right)$

${f}^{'} \left(- 5\right) = - 6 \cdot \left(- 5 + 4\right) \cdot \left(- 5 - 1\right)$

${f}^{'} \left(- 5\right) = - 6 \cdot \left(- 1\right) \cdot \left(- 6\right) = - 36 \to \textcolor{red}{\text{negative}}$

• $\left(- 4 , 1\right)$

${f}^{- 1} = - 6 \cdot \left(- 1 + 4\right) \cdot \left(- 1 - 1\right)$

${f}^{- 1} = - 6 \cdot 3 \cdot \left(- 2\right) = 36 \to \textcolor{g r e e n}{\text{positive}}$

• $\left(1 , \infty\right)$

${f}^{'} \left(2\right) = - 6 \cdot \left(2 + 4\right) \cdot \left(2 - 1\right)$

${f}^{'} \left(2\right) = - 6 \cdot 6 \cdot 1 = - 36 \to \textcolor{red}{\text{negative}}$

So, the first derivative changes sign twice. It first goes from negative to positive around $x = - 4$, and then from positive to negative around $x = 1$.

This means that the function switches from decreaing to increasing around $x = - 4$, so this critical point is a local minimum.

Then the function switches from increasing to decreasing around $x = 1$, so this critical point is a local maximum.

To get the actual coordinates of the local extrema, evaluate the function $f$ in those two points

$f \left(- 4\right) = - 2 {\left(- 4\right)}^{3} - 9 {\left(- 4\right)}^{2} + 24 \left(- 4\right) + 40$

$f \left(- 4\right) = 128 - 144 - 96 + 40 = - 72$

and

$f \left(1\right) = - 2 \cdot {1}^{3} - 9 \cdot {1}^{2} + 24 \cdot 1 + 40$

$f \left(1\right) = - 2 - 9 + 24 + 40 = 53$

$\textcolor{g r e e n}{\left(- 4 \text{,} - 72\right)} \to$ local minimum
$\textcolor{g r e e n}{\left(1 \text{,} 53\right)} \to$ local maximum