Question

f(x)=xlnx
f'(x)=1+lnx critical number: e^-1
f''(x)=1/x

How is this always concave up?

Answer 1

This is not \(\text{always}\) concave up, only when \(x\in(0,\infty)\).

Answer 2

Stupid webwork confusing me....

Answer 3

f(x) is complex for \[x < 0\] so if we're only considering where f(x) is real then it is always concave up.

Answer 4

That's a good observation.

Answer 5

thanks :D