# How do you find the maximum, minimum and inflection points and concavity for the function F(x) = 2x(x-4)^3?

Feb 22, 2018

Local minimum: (1, -54)
Inflection Points: (4, 0) and (2, -32)
This function has no local maximum.
graph{2x(x-4)^3 [-1.47, 6.0, -60.732, 15.714]}

#### Explanation:

Find local extrema

You can find the local extrema (maximum and minimum) by setting the first derivative to zero.

To find the first derivative, directly apply the product rule:
$F ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[2 x\right] \cdot {\left(x - 4\right)}^{3} + 2 x \cdot \frac{d}{\mathrm{dx}} \left[{\left(x - 4\right)}^{3}\right]$

and then the chain rule along with the power rule:
$F ' \left(x\right) = 2 \cdot {\left(x - 4\right)}^{3} + 2 x \cdot 3 \cdot {\left(x - 4\right)}^{2} \cdot \frac{d}{\mathrm{dx}} \left[\left(x - 4\right)\right]$
$F ' \left(x\right) = 2 \cdot {\left(x - 4\right)}^{3} + 6 x \cdot {\left(x - 4\right)}^{2}$

Factor out $\left(x - 4\right)$ and further simplify
$F ' \left(x\right) = {\left(x - 4\right)}^{2} \cdot \left[2 \cdot \left(x - 4\right) + 6 x\right]$
$F ' \left(x\right) = 8 \cdot \left(x - 1\right) \cdot {\left(x - 4\right)}^{2}$

Now set the first derivative to zero and solve for $x$:
$F ' \left(x\right) = 8 \cdot \left(x - 1\right) \cdot {\left(x - 4\right)}^{2} = 0$

By the factor theorem, we have $x = 1$ and $x = 4$.

Verify local extrema

Still, we aren't 100% sure if these x-coordinates corresponds to extrema; nor do we know whether each of them is a local maximum or a local minimum. To be safe (and to avoid potential pitfalls) we need to apply the second derivative test.

Differentiate the first derivative to find $F ' ' \left(x\right)$
$F ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[F ' \left(x\right)\right] = \frac{d}{\mathrm{dx}} \left[8 \cdot \left(x - 1\right) \cdot {\left(x - 4\right)}^{2}\right]$

Again, apply the product rule, the chain rule, and the power rule:
$F ' ' \left(x\right) = 8 \left[1 \cdot {\left(x - 4\right)}^{2} + 2 \cdot \left(x - 1\right) \cdot \left(x - 4\right) \cdot 1\right]$

Factor out $\left(x - 4\right)$ and simplify:
$F ' ' \left(x\right) = 8 \left(x - 4\right) \left[\left(x - 4\right) + 2 \cdot \left(x - 1\right)\right]$
$F ' ' \left(x\right) = 24 \left(x - 2\right) \left(x - 4\right)$

Now plug zeros of the first derivative into the second derivative. Depending on the sign of the value (or the output's relationship with zero, to be precise,) there is going to be a

• Local Maximum if $F ' ' \left(x\right) > 0$
• Local Minimum if $F ' ' \left(x\right) < 0$
• Point of inflection if $F ' ' \left(x\right) = 0$

In this case we have
$F ' ' \left(1\right) = 72$ and therefore a local minimum at $\left(1 , F \left(1\right)\right)$
$F ' ' \left(4\right) = 0$ and therefore an inflection point candidate at $\left(4 , F \left(4\right)\right)$

Find (other) points of inflections

So now we've found a point of inflection. Still, we need to set the second derivative to zero and solve for $x$ to see if there are any other inflection points left behind.

$F ' ' \left(x\right) = 24 \left(x - 2\right) \left(x - 4\right) = 0$

Again, apply the factor theorem, and we find $x = 2$ and $x = 4$ roots of the second derivative.

Verify points of inflections

graph{(x-2)(x-4) [0.391, 5.39, -1.48, 1.02]}
$\textcolor{g r e y}{\text{Graph of "F''(x).}}$

Important: Make sure that the $F ' ' \left(x\right)$ curve crosses rather than touches the $x$-axis at each point of inflection. That is, the corresponding factor (e.g., $\left(x - 2\right)$ for the point $x = 2$) is raised to an odd power (like ${\left(x - a\right)}^{1}$, ${\left(x - a\right)}^{3}$, etc.)) Otherwise the concavity won't change and there's going to be no point of inflection. In this case, both $x = 1$ and $x = 4$ qualifies.

Now evaluate $F$ at each corresponding value of $x$ and finally, you get:
One local minimum at (1, -54) and two
inflection points at (4, 0) and (2, -32).