Well, you are told that b is a positive constant, therefore any value of x that you plug into f''(x) will yield a positive number since (x^2) is always positive.
Therefore f''(x)>0, which means that you have a local minimum
a) if b is a positive constant and x>0 , find all critical points of
f(x) = x-blnx.
b) use the second derivative test to determine whether the function has local maximum or local minimum at each critical point.
a= I calculated f'(x), and I obtained f'(x) = 1-b/x , so its critical value would be b.
For b , I believe its derivative is f''(x)= b/x^2 . I know that the second derivative indicates inflection points , concavity , and also local max and min , but I don't how to apply the second derivative test for this particular case! thanks for any help
You found that the critical value would be Analyze the second derivative for this point to determine if it is positive or negative.
It was given that b must be a positive number. Positive one over a positive number is going to be positive.
Since the second derivative was positive at that point, this determines that the graph is concave up and therefore this point must be a local minimum.
Looks like I'm too slow, Arturo already explained it. =)