Find min/max along with concavity, inflection points, and asymptotes

Question

For the function ƒ (x)= x²/(x−2)²

 

1.Find the vertical and horizontal asymptotes of  ƒ (x).
2.Find the intervals on which ƒ (x)is increasing or decreasing.
3.Find the critical numbers and specify where local maxima and minima of ƒ (x)occur.
4.Find the intervals of concavity and the inflection points of ƒ (x).

 

So far I have f'(x) = -4x/(x-2)³ and f''(x)=8(x+1)/(x−2)⁴ 

 

I think the critical is x= -1 I am not sure what the domain is to determine the endpoint

I think the inflection point is x=-1

Vertical Asymptote x=2

Horizontal Asymptot y=1

 

I need help with the intervals which are decreasing and increasing, and making sure the maxima and minima are correct

8(x+1)(x−2)4

Answer 1

Brittany,

 

Your formulas for f'(x) and f""(x) are correct.  Here's how the 3 functions behave as x increases from -∞ to +∞.

 

f(x) = x²/(x-2)²  =  [ x/(x - 2)]².  For large negative x it is asymptotic to +1 from below, then always decreases gradually as it goes thru zero at x=0, and starts increasing rapidly toward +∞ as it becomes asymptotic to x=2 from the left. From the right of x=2 it decreases rapidly from +∞ and becomes asymptotic to +1 from above as x gets very large positive.

 

f'(x) = [-4x/(x - 2)³].  It shows a zero slope that becomes slightly negative from -∞ to 0, a zero slope at x=0, then an increasing positive slope to +∞ approaching x=2 from the left. On the other side of x=2, the slope start out very negative and approaches zero as x gets very large.

 

f''(x) = [8(x + 1)/(x - 2)⁴]. It is negative for x in (-∞, -1), zero at x= -1, positive in (0, 2) interval, + ∞ at x+ +2, and positive in (2, +∞).

 

Now the answers to your questions:

 

1.  The vertical asymptote occurs at x= +2, and the horizontal one at large absolute values of x, either - or +.

 

2.  f(x) decreasing on (-∞, 0) and (+2, +∞), and increasing on [0, +2).

 

3.  A local and global minimum occurs at x=0.  At the horizontal asymptotes, f is neither a min or a max. At x = +2, f is a global max, but not a local one since f'(2) is never zero, being either very large positive or very large negative as x approaches +2 from both directions.

 

4.  x = 0 is an inflection point as f'(0) = 0.  No others exist.  F(x) is concave downward in (-∞, -1), and concave upward for x in [-1, +2) and (+2, +∞).