Inflections and such

1. Nov 14, 2006

helpm3pl3ase

A) Vertical and horizontal asymptotes
B) Where it increases and decreases
C) Local Min and Local Max
D) Concavity and Inflection Pts

F(x) = x^2\(x-2)^2

A) Vertical Asymptote x = 2? Any Horizontal??
B) I am a little confused because its a fraction? I know you take the derivative and set it to zero. So would the critical # be -4x=0 so x = 4??
C) Local min and Max are confusing me also because of the fraction. I can do it with a linear function
D)?

Any help would be appreciated. Thanks

2. Nov 14, 2006

FunkyDwarf

A.) Correct for the verticle one, as for the horizontal, what happens when x tends to positive and negative infinity?
B.) Find the stationary points and test using the derivative on either side of those points to find the gradient over intervals and thus where its increasing and decreasing.
C.) Doing the above will help you do that. look up the sign test or 2nd derivative test, tho the latter would probably be a bit messy with a function like this
D.) A point of inflection is a change in concavity, ie going from concave down to concave up.

3. Nov 15, 2006

HallsofIvy

Staff Emeritus
If -4x= 0 then x is NOT 4! You aren't, however, asked about critical points. You are asked about where the derivative is positive or negative. It's a good idea to determine where the derivative is 0 because a continuous function can change sign only where it is 0. However, this function, and its derivative, are not always continuous. After you have the derivative, determine where the numerator is 0 (so the derivative is 0) and where the denominator is 0 (so the derivative is not continuous). Check one point in each interval between those points to see whether the derivative is positive or negative there.

Did you really mean "linear function"? A linear function has NO max or min! Finding where the numerator of the derivative is 0 will help you here. Then determine how the derivative changes from + to - or vice versa to determine if it is a max or min.