How do you find the critical points for xlnx and the local max and min?

1 Answer
Jun 27, 2017

The derivative of xlnx is given by the product rule.

y' = 1(lnx) + x(1/x)

y' = lnx + 1

The critical points occur when the derivative equals 0 or is undefined (the latter will only be a critical point if the point is defined in the original function).

0 = lnx + 1

-1 = lnx

e^-1 = x

The derivative is undefined at x = 0, but the function is as well, so we can't count it as a critical point.

Whenever x > e^-1, the derivative is positive, therefore the function is increasing. This signifies that x = e^-1 is an absolute minimum.

Here is a graphical confirmation.

graph{xlnx [-18.02, 18.01, -9.01, 9.01]}

Hopefully this helps!