How do you find the critical numbers for fx x2ln xSolutionA

Solution

Anonymous solved the problem but did not explain the solution at all.

You take the derivative using the product rule - (fg)\'=f\'g+g\'f
Also, note that the derivative of ln x is 1/x and the power rule for differentiation works even for negative powers - (x^n)\' = nx^(n-1)

Thus, [(x^-2)(lnx)]\' = (x^-2)\' ln x + x^-2(ln x)\' =
-2x^-3 ln x + x^-2 (1/x) =
-2x^-3 ln x + x^-3 =
x^-3(-2 ln x + 1)

Thus, the derivative equals 0 if x^-3 = 0 or -2 ln x + 1 = 0

x^-3 does not equal 0 for any x.
-2 ln x + 1 = 0 implies
-2 ln x = -1, or ln x = 1/2 or, exponentiating both sides,
x = e ^ 1/2
This is the only critical value.

Note that the second derivative is negative here, so we have a maximum.
The second derivative is
[x^-3(-2 ln x + 1)]\' =
-3x^-4(-2 ln x + 1) -2/x^-4 =
6x^-4 ln x - 5 x^-4 =
x^-4(6 ln x - 5) =
x^-4(3(2 ln x - 1) - 2)
I factored it this way as we know that 2 ln x - 1 = 0, so x^-4(-2) is clearly negative.

Other things to note -
the function does not exist for nonpositive x.
As x^-2 goes to infinity as x goes to 0, and ln x goes to - infinity as x goes to 0,
x^-2 ln x goes to -infinity as x goes to 0.

As x goes to infinity, we can use l\'Hopital\'s rule to verify that the limit is 0.

Consider x^-2ln x = ln x/x^2

Take the derivative of the numerator and denominator.
We get 1/x/(2x) = 1/(2x^2)
This goes to 0 as x goes to infinity.

Thus, we have a single critical value at x = e^(1/2)

x^-2 ln x = (e^(1/2))^-2* ln(e^(1/2)) =

1/e(1/2) = 1/(2e)

This is the maximum.

The function is positive and has a 0 asymptote as x tends to infinity; as x goes to 0, the function goes to - infinity.