If y = x^2 ln x, what are the points of inflection, concavity and critical points?

Feb 19, 2016

Critical point at $x = \frac{1}{\sqrt{e}}$, concave down on $\left(0 , \frac{1}{e} ^ \left(\text{3/2}\right)\right)$, concave up on $\left(\frac{1}{e} ^ \left(\text{3/2}\right) , + \infty\right)$, point of inflection at $x = \frac{1}{e} ^ \left(\text{3/2}\right)$

Explanation:

Finding critical points:

For the function $f \left(x\right)$, a critical point at $x = c$ where $f \left(c\right)$ exists is a point where either $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ doesn't exist.

Thus, to find critical values, we must find the derivative of the function. To do this to $y = {x}^{2} \ln x$, we must use the product rule.

$y ' = \ln x \frac{d}{\mathrm{dx}} \left({x}^{2}\right) + {x}^{2} \frac{d}{\mathrm{dx}} \left(\ln x\right)$

$y ' = \ln x \left(2 x\right) + {x}^{2} \left(\frac{1}{x}\right)$

$y ' = 2 x \ln x + x$

The critical points are where $y ' = 0$ or where $y '$ is undefined where $y$ wasn't.

$2 x \ln x + x = 0$

$x \left(2 \ln x + 1\right) = 0$

This can be split into two equations equalling $0$:

$x = 0$

This potential critical point is discarded since $y '$ doesn't exist at $x = 0$.

$2 \ln x + 1 = 0$

$\ln x = - \frac{1}{2}$

$x = {e}^{- \text{1/2}} = \frac{1}{\sqrt{e}}$

This is the only critical value: $\textcolor{red}{x = \frac{1}{\sqrt{e}}}$

Finding concavity and points of inflection:

Concavity, convexity, and points of inflection are all dictated by a function's second derivative.

• $y$ is concave upwards (convex) when $y ' ' > 0$.
• $y$ has a point of inflection when $y ' ' = 0$ and the concavity shifts
• $y$ is concave downwards (concave) when $y ' ' < 0$.

To find the function's second derivative, again use the power rule.

$y ' = 2 x \ln x + x$

$y ' ' = \ln x \frac{d}{\mathrm{dx}} \left(2 x\right) + 2 x \frac{d}{\mathrm{dx}} \left(\ln x\right) + 1$

$y ' ' = \ln x \left(2\right) + 2 x \left(\frac{1}{x}\right) + 1$

$y ' ' = 2 \ln x + 2 + 1$

$y ' ' = 2 \ln x + 3$

To determine when this is $< 0 , > 0 ,$ or $= 0$, first find when it equals just $0$. This could be a spot where the concavity shifts (the sign changes from positive to negative, or vice versa). If the concavity shifts, there is a point of inflection at that point.

$2 \ln x + 3 = 0$

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

$x = {e}^{- \text{3/2}}$

$x = \frac{1}{e} ^ \left(\text{3/2}\right)$

This is a possible point of inflection. We should test the concavity (sign of the second derivative) around the point $x = \frac{1}{e} ^ \left(\text{3/2}\right)$.

Our test points surrounding $x = \frac{1}{e} ^ \left(\text{3/2}\right)$ can be $x = 0.1$ and $x = 1$, since $\frac{1}{e} ^ \left(\text{3/2}\right) \approx 0.2231$.

At $x = 0.1 :$

$y ' ' = 2 \ln \left(0.1\right) + 3 \approx - 1.605$

At $x = 1 :$

$y ' ' = 2 \ln \left(1\right) + 3 = 3$

Since the second derivative is negative at $x = 0.1$, we know that the entire interval $\left(0 , \frac{1}{e} ^ \left(\text{3/2}\right)\right)$ is concave down.

Since it's positive at $x = 1$, we know the entire interval $\left(\frac{1}{e} ^ \left(\text{3/2}\right) , + \infty\right)$ is concave up.

Since the concavity shifts at $x = \frac{1}{e} ^ \left(\text{3/2}\right)$, we know there is a point of inflection there.