Question

How do you find the critical numbers for `f(x) = x^(-2)ln(x)` to determine the maximum and minimum?

Answer 1

`0` is the minimum

`e^(1/2)` is the maximum

Explanation:

Take the derivative of `f(x)`. You will need to use the product rule. You also need to know that the derivative of `ln(x)` is `1/x`:

`f'(x)=x^-2(1/x) + ln(x)(-2x^-3)`

`f'(x)=x^-3-2ln(x)x^-3`

Factor out a `x^-3`

`f'(x)=x^-3(1-2ln(x))`

Solve for `x`:

`x=0,e^(1/2)`

Plug in these numbers into the initial equation:

`f(0)=0ln(0)=DNE`.

We'll need to use limits for this:

`lim_(xrarr0) x^-2ln(x)=-oo`. This is definitely a minimum

`f(e^(1/2))=(e^-1)(ln(e^(1/2)))=1/(2e)`. This is a maximum because plugging in anything before or after this will give a value less than this.