Question

Graph: f(x)=x-xlnx, f'(x)= -lnx, and f"(x)= -1/x. I am not sure what to do for critical points. And it looks like there are a few holes in my function.

Answer 1

For starters, you know that the ln(x) is undefined for negative values and 0. Second, you know that the -ln(x) = 0 at x=1; which means that the function will begin decreasing after that(the ln(x) negative for values between (0 and 1), making -ln(x) positive.

So you know your function is increasing from (0,1) and decreasing on (1, infinity)

For convacity, -1/x is negative for all positive values of x, which means your function is concave down? I've forgotten the terminology; basically, it looks like a helmet, or has the same concavity as -x^2

Does that answer your question?

Answer 2

Actually oh man. Finally my brain is working. So realizing that logs can't equal a neg value it means my domain is [0, infinity). and I also realize I have a max at 1. Where the slope goes from inc to decreasing. Yeah I think I've got it.