# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for f(x) = x ln x?

Jan 26, 2017

$f \left(x\right)$ is decreasing on $\left(0 , \frac{1}{e}\right)$ and increasing on $\left(\frac{1}{e} , \infty\right)$. Therefore, this point is a relative minimum. There are no relative maxima.

#### Explanation:

In order to find the relative extrema (maxima or minima) for a function, we must find its derivative. This is true because a function is increasing when its derivative is positive and decreasing when its derivative is negative.

However, before we do this, a step students often forget is to consider the domain of the function. Relative extrema cannot exist at points where the function itself does not exist! In this case, $\ln x$ does not exist when $x \le 0$. Therefore, the domain of $f \left(x\right)$ is $\left(0 , \infty\right)$.

We now find the derivative using the product rule, since $f \left(x\right)$ is a product of two functions:

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

We now consider the "critical points" (points in the domain of $f \left(x\right)$ where $f ' \left(x\right) = 0$ or does not exist). These would be places where $f '$ could change signs, causing $f$ to change directions.

In the domain of $\left(0 , \infty\right)$:
there are no points where $f '$ does not exist.
We now look for points in the domain where $f ' \left(x\right) = 0$.
$f ' \left(x\right) = 0$
$1 + \ln x = 0$
$\ln x = - 1$
$x = {e}^{- 1} = \frac{1}{e} = 0$
We now apply the first derivative test. If we select a number in the domain smaller than $x \frac{1}{e} = 0.36787944 . . .$ such as $x = 0.1$,
$f ' \left(0.1\right) = 1 + \ln \left(0.1\right) = - 1.30258 . . .$.
If we select a number in the domain larger than $x = \frac{1}{e}$ such as $x = 5$,
$f ' \left(5\right) = 1 + \ln \left(5\right) = 2.68943 . . .$.

Since the sign of $f ' \left(x\right)$ changes from negative to positive at $x = \frac{1}{e}$, the direction of $f \left(x\right)$ changes from decreasing to increasing at this point, making it a relative minimum by the first derivative test. In fact, since it is the only critical number, it is also the absolute minimum.