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

Jan 17, 2016

Increasing.

#### Explanation:

The derivative of a function can be used to determine if a function is increasing or decreasing at a point.

• If $f ' \left(1\right) < 0$, then $f \left(x\right)$ is decreasing at $x = 1$.
• If $f ' \left(1\right) > 0$, then $f \left(x\right)$ is increasing at $x = 1$.

First, find $f ' \left(x\right)$ through the product rule.

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

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

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

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

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

$f ' \left(1\right) = 2 \left(0\right) + 1$

$f ' \left(1\right) = 1$

Since $f ' \left(1\right) > 0$, the function is increasing at $x = 1$.

We can check a graph of $f \left(x\right)$:

graph{x^2lnx [-0.221, 1.6745, -0.353, 0.595]}