Question

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`?

Answer 1

`f(x)` is decreasing on `(0,1/e)` and increasing on `(1/e, oo)`. 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, `lnx` does not exist when `x<=0`. Therefore, the domain of `f(x)` is `(0,oo)`.

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

`f(x)=xlnx`
`f'(x)=x*(1/x) +lnx*1`
`f'(x)=1+lnx`

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

In the domain of `(0,oo)`:
there are no points where `f'` does not exist.
We now look for points in the domain where `f'(x)=0`.
`f'(x)=0`
`1+lnx=0`
`lnx=-1`
`x=e^(-1)=1/e=0`
We now apply the first derivative test. If we select a number in the domain smaller than `x1/e=0.36787944 . . .` such as `x=0.1`,
`f'(0.1) =1+ln(0.1)=-1.30258 . . .`.
If we select a number in the domain larger than `x=1/e` such as `x=5`,
`f'(5) =1+ln(5)=2.68943 . . .`.

Since the sign of `f'(x)` changes from negative to positive at `x=1/e`, the direction of `f(x)` 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.