# How do use the first derivative test to determine the local extrema f(x) = 3x^5 - 20x^3?

Jul 27, 2015

You determine the crititcal points of the function and check to see if any of these points are local minima or local maxima.

#### Explanation:

Basically, the first derivative test allows you to look for a function's critical points and check to see whether or not these points are local extrema by looking if the sign of the first derivative changes around them.

So, the first thing you need to do is determine the first derivative of $f$, which is

${f}^{'} = 15 {x}^{4} - 60 {x}^{2}$

To find the function's critical points, make the first derivative equal to zero and solve for $x$

${f}^{'} = 0$

$15 {x}^{4} - 60 {x}^{2} = 0$

$15 {x}^{2} \left({x}^{2} - 4\right) = 0$

$15 {x}^{2} \cdot \left(x - 2\right) \left(x + 2\right) = 0$

This equation has three solutions, $x = 0$ and $x = \pm 2$.

Since your function has no domain restrictions, all these points will be critical points.

Now, in order for the points to be local extrema, the function must go from increasing, i.e. a positive ${f}^{'}$, to decreasing, i.e. a negative ${f}^{'}$, or vice versa.

Since you have 3 critical points, you're going to look at 4 intervals. Select a value from each itnerval to determine the sign of the first derivative on that interval

• $\left(- \infty , - 2\right)$

${f}^{'} \left(- 3\right) = 15 \cdot {\left(- 3\right)}^{4} - 60 \cdot {\left(- 3\right)}^{2}$

${f}^{'} \left(- 3\right) = 1215 - 540 = 675$ $\to$ $\textcolor{g r e e n}{\text{positive}}$

• $\left(- 2 , 0\right)$

${f}^{'} \left(- 1\right) = 15 \cdot {\left(- 1\right)}^{4} - 60 \cdot {\left(- 1\right)}^{2}$

${f}^{'} \left(- 1\right) = 15 - 60 = - 45$ $\to$ $\textcolor{red}{\text{negative}}$

• $\left(0 , 2\right)$

${f}^{'} \left(1\right) = 15 \cdot {\left(1\right)}^{4} - 60 \cdot {\left(1\right)}^{4}$

${f}^{'} \left(1\right) = 15 - 60 = - 45 \to \textcolor{red}{\text{negative}}$

• $\left(2 , 0\right)$

${f}^{'} \left(3\right) = 15 \cdot {\left(3\right)}^{4} - 60 \cdot {\left(3\right)}^{2}$

${f}^{'} \left(3\right) = 1215 - 540 = 675 \to \textcolor{g r e e n}{\text{positive}}$

So, you know that the first derivative changes sign, i.e. it goes from positive to negative, around $- 2$, so this point is a local maximum.

On the other hand, the first derivative does not change sign around point $0$, so this point will not be of interest.

Notice that the first derivative changes sign again around point $+ 2$, since it goes from negative to positive. This means that point $2$ will be a local minimum.

To get the actual points that match these critical points, evaluate the original functions in $x = - 2$ and $x = 2$

$f \left(- 2\right) = 3 \cdot {\left(- 2\right)}^{5} - 20 \cdot {\left(- 2\right)}^{3}$

$f \left(- 2\right) = - 96 + 160 = 64$

and

$f \left(2\right) = 3 \cdot {\left(2\right)}^{5} - 20 \cdot {\left(2\right)}^{3}$

$f \left(2\right) = 96 - 160 = - 64$

Therefore, the function $f = 3 {x}^{5} - 20 {x}^{3}$ has a local minimum at $\textcolor{g r e e n}{\left(2 \text{,} - 64\right)}$ and a local maximum at $\textcolor{g r e e n}{\left(- 2 \text{, } 64\right)}$.