# What are the relative extrema of this equation? x^4 - 2x^3

##### 2 Answers
Feb 26, 2018

Local minimum at $\left(\frac{3}{2} , - \frac{27}{16}\right)$

#### Explanation:

To find local extrema, we use the first an second derivative tests.

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

For the sake of the first derivative test, let's factor this to:

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

If we set $f ' \left(x\right) = 0$, we find two possible extrema at $x = 0$ and $x = \frac{3}{2}$

Now we use the second derivative test to determine minimum/maximum/point of inflection:

$f ' \left(x\right) = 4 {x}^{3} - 6 {x}^{2}$
$f ' ' \left(x\right) = 12 {x}^{2} - 12 x$

Factoring, we get

$f ' ' \left(x\right) = 12 x \left(x - 1\right)$

Now we evaluate our two possible extrema using the second derivative test:

$f ' ' \left(0\right) = 12 \left(0\right) \left(0 - 1\right) = 0$ This is a point of inflection, not an extreme.

$f ' ' \left(\frac{3}{2}\right) = 12 \left(\frac{3}{2}\right) \left(\frac{3}{2} - 1\right) = 9$ The graph is concave up at $x = \frac{3}{2}$ and is therefore a local minimum. Plugging this $x$ back into our original function gives a local minimum at point $\left(\frac{3}{2} , - \frac{27}{16}\right)$

All of this can be observed on the graph of the original function:

graph{x^4-2x^3 [-1.49, 4.67, -1.958, 1.122]}

Feb 26, 2018

We have a local minimum at $\left(\frac{3}{2} , - \frac{27}{16}\right)$

#### Explanation:

Determine $f ' \left(x\right)$, set $f ' \left(x\right) = 0$, solve for $x$ and determine the values of $x$ for which $f ' \left(x\right)$ doesn't exist:

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

$4 {x}^{3} - 6 {x}^{2} = 0$

$2 {x}^{2} \left(2 x - 3\right) = 0$

$2 {x}^{2} = 0 :$

${x}^{2} = 0$
$x = 0$

$2 x - 3 = 0 :$

$2 x = 3$
$x = \frac{3}{2}$

$f ' \left(x\right)$ is a polynomial; therefore, it exists for all real $x$.

We have prospective extrema at $x = 0 , x = \frac{3}{2}$. Let's split up the domain of $f \left(x\right)$, which is (-∞,∞), across these $x$-values. This gives us the intervals:

(-∞,0),(0,3/2),(3/2,∞)

At each of these intervals, we want to determine if $f ' \left(x\right)$ is positive or negative. We'll do this by selecting a value for $x$ from each interval and plugging it into $f ' \left(x\right)$. If $f ' \left(x\right) > 0$ on an interval, $f \left(x\right)$ is increasing on that interval. If $f ' \left(x\right) < 0$ on an interval, $f \left(x\right)$ is decreasing on that interval.

We have an extremum at any $x$ value where $f ' \left(x\right)$ changes signs, IE, $f \left(x\right)$ changes from increasing to decreasing (or vice versa).

(-∞,0):

Let's select $x = - 1$.

$f ' \left(- 1\right) = 4 {\left(- 1\right)}^{3} - 6 \left(- {1}^{2}\right) = - 4 - 6 < 0$

$f \left(x\right)$ is decreasing on (-∞,0)

$\left(0 , \frac{3}{2}\right) :$

Let's select $x = 1.$

$f ' \left(1\right) = 4 - 6 < 0$

$f \left(x\right)$ is also decreasing on $\left(0 , \frac{3}{2}\right) .$

(3/2, ∞):

Let's select $x = 2.$

$f ' \left(2\right) = 4 \left({2}^{3}\right) - 6 \left({2}^{2}\right) = 4 \left(8\right) - 6 \left(4\right) = 32 - 24 > 0$

$f \left(x\right)$ is increasing on (3/2, ∞). This is a switch from the decrease on the previous two intervals, so we have an extremum at $x = \frac{3}{2}$. Since $f \left(x\right)$ went from decreasing on $\left(0 , \frac{3}{2}\right)$ to increasing on (3/2,∞), we have a local minimum at $x = \frac{3}{2.}$

To determine the $y$-coordinate of our local minimum, find $f \left(\frac{3}{2}\right) :$

$f \left(\frac{3}{2}\right) = {\left(\frac{3}{2}\right)}^{4} - 2 {\left(\frac{3}{8}\right)}^{3} = \frac{81}{16} - 2 \left(\frac{27}{8}\right) = \frac{81}{16} - \frac{54}{8} = \frac{81}{16} - \frac{108}{16} = - \frac{27}{16}$

We have a local minimum at $\left(\frac{3}{2} , - \frac{27}{16}\right)$