SOLUTION: Find the absolute maximum and absolute minimum values of the function below. If an absolute maximum or minimum does not exist, enter NONE. f(x) = x^2 + 432/x on the open interval

Algebra ->  Absolute-value -> SOLUTION: Find the absolute maximum and absolute minimum values of the function below. If an absolute maximum or minimum does not exist, enter NONE. f(x) = x^2 + 432/x on the open interval       Log On


   

Question 361533: Find the absolute maximum and absolute minimum values of the function below. If an absolute maximum or minimum does not exist, enter NONE.
f(x) = x^2 + 432/x on the open interval (0, infinity)
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
An absolute minimum exists at x = 6.
y+=+x%5E2+%2B+432%2Fx,
dy%2Fdx=+2x-432%2Fx%5E2. Letting this derivative equal 0, and solving,
we get 2x+=+432%2Fx%5E2, or x%5E3+=+216, or x = 6.
Now d%5E2y%2Fdx%5E2+=+2%2B864%2Fx%5E3. Substitute x = 6 into the 2nd derivative to find out that d%5E2y%2Fdx%5E2%3E0. Therefore there is an absolute minimum at x = 6.