Question

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

Answer 1

`f(x)` is neither increasing nor decreasing when `x=1`.

Explanation:

Find `f(1)`.

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

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

`f'(x)=g'(x)h(x)+h'(x)g(x)`

First, note that I'm assuming you meant `(ln(x))^2`, and not `ln(x^2)`.

`f'(x)=(ln(x))^2d/dx(x)+xd/dx((ln(x))^2)`

Find each derivative:

`d/dx(x)=1`

The next requires chain rule:

`d/dx((ln(x))^2)=2ln(x)d/dx(ln(x))=2ln(x)(1/x)`

Plug these back in to find `f'(x)`.

`f'(x)=(ln(x))^2(1)+x(2ln(x)(1/x))`

Simplify.

`f'(x)=(ln(x))^2+2ln(x)`

Now, find `f'(1)`.

`f'(1)=(ln(1))^2+2ln(1)=0+0=0`

Thus, `f(x)` is neither increasing nor decreasing when `x=1`.

This is the graph of `f(x)`:

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

There is a minimum when `x=1`.