# Maxima and Minima

• June 4th 2009, 10:57 AM
lisa1984wilson
Maxima and Minima
I need to find the local maxima and minima using the second derivative of X^2/x-2.

Now I know how to do this with the first derivative which is (x-4)x/(x-2)^2

(0,0) Local max and (4,8) for local min

derivative 2 equals 8/(x-2)^3

I'm lost with this one!
• June 4th 2009, 11:06 AM
Twig
You have the derivatives, both the second and the first.

Now, if the second derivative is negative for the x-value at which the function attains an extremevalue, then you have a maxima.

If the second derivative is positive, you have a minima.

So you evaluate the second derivative for $x=0 \mbox{ and } x=4$ .

This is sometimes called the second-derivative test.
• June 4th 2009, 11:10 AM
Rapha
Hi!

Quote:

Originally Posted by lisa1984wilson
I need to find the local maxima and minima using the second derivative of X^2/x-2.

It is f(x) = x^2/(x-2).

You find the critical points by solving f'(x) = 0

the second derivative tells you something about min or max.

Quote:

Originally Posted by lisa1984wilson
Now I know how to do this with the first derivative which is (x-4)x/(x-2)^2

(0,0) Local max and (4,8) for local min

derivative 2 equals 8/(x-2)^3

I'm lost with this one!

Why is that? Everything is correct.

f''(x) = 8/(x - 2)^3

f''(0) = -1 < 0 => (0,0) is maximum

f''(4) = 1 > 0 => (4,8) is minimum

Yours
Rapha

Edit: Someone was faster than me.