# Math Help - determine local max or local min

1. ## determine local max or local min

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

2. 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

3. Originally Posted by vance
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 $x = b.$ Analyze the second derivative for this point to determine if it is positive or negative.

$f''(b) = \frac{b}{b^2} = \frac{1}{b}$ 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. =)

4. now, I understand it !!! thanks