# How do you find all relative extrema and points of inflection for the equation: y=x^2 log_3x?

Jul 5, 2015

Find extrema using the derivative and points of inflection using the second derivative.

#### Explanation:

We will make use of: ${\log}_{3} x = \ln \frac{x}{\ln} 3$ and

$\frac{d}{\mathrm{dx}} {\log}_{3} x = \frac{d}{\mathrm{dx}} \left(\ln \frac{x}{\ln} 3\right) = \frac{1}{x \ln 3}$

Finding relative extrema

$f \left(x\right) = y = {x}^{2} {\log}_{3} x$

Note that the domain of this function is $\left(0 , \infty\right)$

Use the product rule to get:

$y ' = 2 x {\log}_{3} x + {x}^{2} \cdot \frac{1}{x \ln 3}$

$= 2 x {\log}_{3} x + \frac{x}{\ln} 3$

To find critical numbers, it is convenient to use all logarithms with the same base.:

$f ' \left(x\right) = y ' = \frac{2 x \ln x}{\ln} 3 + \frac{x}{\ln} 3 = \frac{x \left(2 \ln x + 1\right)}{\ln} 3$

$y '$ in never undefined in the domain, and is equal to $0$ at

$x = 0$ or $2 \ln x + 1 = 0$. $0$ is not in the domain, so we need only solve:
$2 \ln x + 1 = 0$.
$\ln x = - \frac{1}{2}$

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

The partitions on the domain are: $\left(0 , {e}^{- \frac{1}{2}}\right) , \left({e}^{- \frac{1}{2}} , \infty\right)$

Testing $\left(0 , {e}^{- \frac{1}{2}}\right)$.
Note that $\frac{1}{e} < \frac{1}{\sqrt{e}}$, so ${e}^{-} 1$ is in the first interval and

$f ' \left({e}^{- 1}\right) = \frac{{e}^{- 1} \left(2 \ln {e}^{- 1} + 1\right)}{\ln} 3 = \frac{\frac{1}{e} \left(- 2 + 1\right)}{\ln} 3$ which is negative.

So $f '$ is negative on $\left(0 , {e}^{- \frac{1}{2}}\right)$.

Testing $\left({e}^{- \frac{1}{2}} , \infty\right)$
Use $1$ as the test number.

$f ' \left(e\right) = \frac{1 \left(2 \ln 1 + 1\right)}{\ln} 3 = \frac{1 \left(0 + 1\right)}{\ln} 3$ which is positive, so

So $f '$ is positive on $\left({e}^{- \frac{1}{2}} , \infty\right)$.

This tells us that f(e^(-1/2)) is a relative minimum.

Finding points of inflection

Recall:

$f ' \left(x\right) = y ' =$ = 2xlog_3x + x/ln3

So

$f ' ' \left(x\right) = y ' ' = 2 {\log}_{3} x + 2 x \cdot \frac{1}{x \ln 3} + \frac{1}{\ln} 3$

$= \frac{2 \ln x}{\ln} 3 + \frac{2}{\ln} 3 + \frac{1}{\ln} 3 = \frac{2 \ln x + 3}{\ln} 3$

The only partition number for $f ' '$ is $\ln x = - \frac{3}{2}$ so $x = {e}^{- \frac{3}{2}}$

For $f ' '$ the parition intervals are $\left(0 , {e}^{- \frac{3}{2}}\right) , \left({e}^{- \frac{3}{2}} , \infty\right)$

Using ${e}^{-} 2$ and $1$ as test numbers we will find that:

$f ' '$ is negative on $\left(0 , {e}^{- \frac{3}{2}}\right)$.

$f '$ is positive on $\left({e}^{- \frac{3}{2}} , \infty\right)$.

The concavity does change, so

$\left({e}^{- \frac{3}{2}} , f \left({e}^{- \frac{3}{2}}\right)\right)$ is a point of inflection.