# Is f(x)=xln(x)^2 increasing or decreasing at x=1?

Jan 7, 2016

$f \left(x\right)$ is neither increasing nor decreasing when $x = 1$.

#### Explanation:

Find $f \left(1\right)$.

If $f \left(1\right) > 0$, the function is increasing when $x = 1$.
If $f \left(1\right) < 0$, the function is increasing when $x = 1$.
If $f \left(1\right) = 0$, there is a critical value and the function is neither increasing nor decreasing.

To find $f \left(x\right)$, use the product rule, which states that for a function $f \left(x\right) = g \left(x\right) h \left(x\right)$, then

$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + h ' \left(x\right) g \left(x\right)$

First, note that I'm assuming you meant ${\left(\ln \left(x\right)\right)}^{2}$, and not $\ln \left({x}^{2}\right)$.

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

Find each derivative:

$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

The next requires chain rule:

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

Plug these back in to find $f ' \left(x\right)$.

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

Simplify.

$f ' \left(x\right) = {\left(\ln \left(x\right)\right)}^{2} + 2 \ln \left(x\right)$

Now, find $f ' \left(1\right)$.

$f ' \left(1\right) = {\left(\ln \left(1\right)\right)}^{2} + 2 \ln \left(1\right) = 0 + 0 = 0$

Thus, $f \left(x\right)$ is neither increasing nor decreasing when $x = 1$.

This is the graph of $f \left(x\right)$:

graph{x(ln(x))^2 [-3.16, 10.887, -1.46, 5.565]}

There is a minimum when $x = 1$.