Question

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

Answer 1

Increasing.

Explanation:

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

• If `f'(1)<0`, then `f(x)` is decreasing at `x=1`.
• If `f'(1)>0`, then `f(x)` is increasing at `x=1`.

First, find `f'(x)` through the product rule.

`f'(x)=lnxd/dx[x^2]+x^2d/dx[lnx]`

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

`f'(x)=2xlnx+x`

Now, find `f'(1)`.

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

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

`f'(1)=1`

Since `f'(1)>0`, the function is increasing at `x=1`.

We can check a graph of `f(x)`:

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