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

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Answer 1 · 2017-04-15T20:13:18

$0$ is the minimum

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

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.

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