# How do you find points of inflection and determine the intervals of concavity given y=4x^2e^(3x)?

Sep 27, 2017

Use the second derivative $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ to find possible points of inflection, and then test those by evaluating the second derivative using test points in intervals around those points.

#### Explanation:

First find $\frac{\mathrm{dy}}{\mathrm{dx}}$ using the Chain Rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 {x}^{2}\right) \left(3 \cdot {e}^{3 x}\right) + \left({e}^{3 x}\right) \left(2 \cdot 4 x\right)$

$= 12 {x}^{2} {e}^{3 x} + 8 x {e}^{3 x}$
$= 4 x {e}^{3 x} \left(3 x + 2\right)$

Next, find $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ by differentiating again.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left(4 x {e}^{3 x}\right) \left(3\right) + \left(3 x + 2\right) \left[\left(4 x\right) \left(3 \cdot {e}^{3 x}\right) + \left({e}^{3 x} \cdot 4\right)\right]$
$= 12 x {e}^{3 x} + \left(3 x + 2\right) \left(12 x {e}^{3 x} + 4 {e}^{3 x}\right)$
$= 12 x {e}^{3 x} + 36 {x}^{2} {e}^{3 x} + 12 x {e}^{3 x} + 24 x {e}^{3 x} + 8 {e}^{3 x}$
$= 36 {x}^{2} {e}^{3 x} + 48 x {e}^{3 x} + 8 {e}^{3 x}$
$= 4 {e}^{3 x} \left(9 {x}^{2} + 12 x + 2\right)$

Potential points of inflection will occur at values of $x$ where the second derivative is 0 or undefined. By examining $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ as derived, there are no places where it is undefined. To find places where it is 0, we set $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$ and solve:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$
$4 {e}^{3 x} \left(9 {x}^{2} + 12 x + 2\right) = 0 \implies 4 {e}^{3 x} = 0 \mathmr{and} \left(9 {x}^{2} + 12 x + 2\right) = 0$

$4 {e}^{3 x} = 0$ has no solutions.

$9 {x}^{2} + 12 x + 2 = 0$ requires the Quadratic Formula:

$x = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \cdot 9 \cdot 2}}{2 \cdot 9} = \frac{- 12 \pm \sqrt{144 - 72}}{18}$
$= \frac{- 12 \pm \sqrt{72}}{18} = \frac{- 12 \pm 6 \sqrt{2}}{18} = \frac{- 2 \pm \sqrt{2}}{3}$

This results in values of $x \approx - 0.195 , - 1.1 .38$

To determine if either are points of inflection, and to determine intervals of concavity, test values of $x$ that "surround" these possible points of inflection, paying close attention to the sign of the second derivative at each test point. We can use $x$ values of -2, -1, and 0. Note that $4 {e}^{3 x}$ is always positive, and therefore the sign of the second derivative depends solely upon the second quadratic term:

$x = - 2 \implies 9 {\left(- 2\right)}^{2} + 12 \left(- 2\right) + 2 = 36 - 24 + 2 = 62 > 0$
$x = - 1 \implies 9 {\left(- 1\right)}^{2} + 12 \left(- 1\right) + 2 = 9 - 12 + 2 = - 1 < 0$
$x = 0 \implies 9 {\left(0\right)}^{2} + 12 \left(0\right) + 2 = 0 + 0 + 2 = 2 > 0$

If the second derivative is positive, the function is concave up in that interval. Thus, in the intervals $\left(- \infty , \frac{- 2 - \sqrt{2}}{3}\right)$ and $\left(\frac{- 2 + \sqrt{2}}{3} , \infty\right)$ the function is concave up, while in the interval $\left(\frac{- 2 - \sqrt{2}}{3} , \frac{- 2 + \sqrt{2}}{3}\right)$ the function is concave down.

Since each value for $x$ determined above is a place where the concavity changes, each of those values of $x$ are points of inflection. This can be verified by examining the graph of the function:

graph{4x^2e^(3x) [-2.496, 0.542, -0.557, 0.962]}