Question

For the function below, determine the intervals of concavity:
f(x)=xlnx
I tried doing it but the second derivative would always be f''(x)=1/x, so it would be concave up at x>0 and concave down at x<0, which is wrong.

Also, for the function below, determine the intervals of increase:
f(x)=2^(x/3)
I can't seem to isolate the x to find the critical points?

Find all the critical points and determine whether each is a local maximum or minimum or neither:
f'(x)=x(9-x^2)^(1/2)
I was just wondering if the maxes are at -3 and 3?

Answer 1

while f" = 1/x, f is not defined for x<=0. So, f is concave up for all x>0.

f = 2^(x/2)
f' = ln2/2 2^(x/2)
f" = (ln2/2)^2 2^(x/2)
This is just e^kx in disguise. It is concave up everywhere. There are no critical points.

f' = x√(9-x^2)
f'=0 at x=0, x=±3
f" = (9-2x^2)/√(9-x^2)
Since f">0 at x=0, that is a minimum
Since f" is undefined at x=±3, neither min nor max.

play around with these functions at wolframalpha.com to see the graphs.

Answer 2

1st question:
The answer is just that
"The graph is concave upward when x > 0"
Your approach to find the second derivative is correct and your second derivative is correct.
However, you should note that when x<0,
ln(x) is undefined, i.e. the original function is undefined when x <0 and so as the second derivative. (Domain of the function: x>0
That means you should not consider the concavity for x < 0 and that is the wrong thing about your answer.