Find extreme values!

Question

y = 2/(x-2)^2 - (x). Find extreme values for this function (if there are any) and draw the function.

Answer 1

When you're looking for extreme values of a function, you need to check three things: critical values (where the derivative is zero), discontinuities, and end conditions.

 

Lets start with end conditions.

What is the limit of this function as x->-∞ and x->+∞?

The first term has a limit of 0 in both cases, so y->+∞ as x->-∞, and y->-∞ as x->+∞. Therefore, you have no absolute minimum or maximum.

 

This function has one discontinuity at x=2. If you take the limit as x->2, you'll find that y->+∞ on both sides. The denominator is squared, so the first term is always positive, and subtracting 2 from +∞ does nothing.

 

Now lets find any critical values.

y = 2(x-2)^-2 - x

y' = -4(x-2)^-3 - 1 = 0

... and you can solve for x. To determine if this is a local minimum, maximum, or neither we look at the 2nd derivative.

 

y'' = 12(x-2)^-4

This is always positive (even power) on its domain, so the function is concave up everywhere except x=2, and the critical value found earlier must be a minimum.

 

To sketch, put all these pieces together. The end conditions indicate that the function behaves as y= -x for large x values. There is an asymptote at x=2, and the function goes toward positive infinity on both sides. There is a local minimum to the left of 2 at the critical value found earlier. You may want to find the y-value of this point for a more accurate graph. The graph is also concave up everywhere on its domain, so make sure it always curls in the proper direction.

 

Hope this helps, please comment if you have further questions!