# Critical Points- Does this function not have any?

• Apr 18th 2008, 12:43 PM
NAPA55
Critical Points- Does this function not have any?
Does the function:

f(x) = 3 + (1/(x+2))

Have any critical (relative max/min) points?

My answer is NO, but I'm not 100% certain this is correct.

The method we were taught is to find the derivative of the function, set it equal to zero, and find the x values. Then we look at intervals to the right and left of this point, and if they are different (pos to neg or neg to pos), then it is a max or min, respectively.

Finding the derivative of this function and setting it equal to zero, then solving for x, yields a result of 1. Finding the derivative of values left and right of this point gives 2 positive answers, meaning that 1 is NOT a critical point, and therefore there are no critical points.

The graph has a vertical asymptote at -2 as the function approaches from the left and the right.

The graph has a horizontal asymptote at 3 as well.

Both of these conditions seem to imply that there are no maximum or minimum points.

Yes? No?
• Apr 18th 2008, 01:09 PM
colby2152
Quote:

Originally Posted by NAPA55
Does the function:

f(x) = 3 + (1/(x+2))

Have any critical (relative max/min) points?

My answer is NO, but I'm not 100% certain this is correct.

$3+\frac{1}{(x+2)^2}$

The derivative is:

$\frac{-2}{(x+2)^3}$

Set it equal to zero:

$\frac{-2}{(x+2)^3} = 0$

You cannot solve for x, hence there being no critical points.
• Apr 18th 2008, 01:36 PM
topsquark
Quote:

Originally Posted by NAPA55
f(x) = 3 + (1/(x+2))

There is a kind of critical point at x = -2. Critical points are also where the function or first derivative don't exist. A point where the function exists, but the first derivative does not can be either a max or a min.

-Dan
• Apr 18th 2008, 01:56 PM
NAPA55
Interesting... thank you both! (Clapping)