Consider the function f(x) =x^2 / x^2 − 4. Find the vertical and horizontal asymptotes.

Question

Hi, looking for help on the following problem that has multiple parts: please show the work

 

Consider the function f(x) =x^2/x^2 − 4.

A-Find the vertical and horizontal asymptotes.

B-Find the interval where the function is increasing. Find the interval where the function is decreasing.

C-Find the local maximum value.

D-ind the interval where the function is concave up. Find the interval where the function is concave down.

 

Thanks so much,

Stephanie

Answer 1

f(x) =x²/(x²-4)

A)Vertical asymptotes are: x=±2.

lim_x→2⁺f(x)=lim_x→-2^-f(x)=+∞

lim_x→2^-f(x)=lim_x→-2⁺f(x)= -∞.

To find horizontal asymptote(s), we rewrite the function in the form

f(x)=1+4/(x²-4).

lim_x→±∞f(x)=1. So y=1 is a horizontal asymptote.

(Conditionally, y=0 can be considered as a horizontal asymptote for -2<x<2 as well, but principally it is not an asymptote.)

 

B. f(x)=1+4/(x²-4)

f'(x)=-8x/(x²-4)³

From here it follows that x=0 is a critical point and this gives a local maximum because f''(0)=-1/2 <0 (you have to show it yourself!)

Next, we have two other critical points x=-2 and x=2.

Now we have to clarify how f(x) behaves (decreases or increases) in each of the intervals (-∞, -2), (-2, 0), (0,2), (2, +∞). An easy way to see it is to determine the sign of the f'(x) in each interval (enough to show for one point!).

 

Interval |(-∞, -2)| -2| (-2, 0)|0|(0, 2)| 2  |(2, +∞)|

f'(x)      |     +     |na |     +   |0|   -    |na |     -    |

f(x)       |     ↑     |±∞|     ↑    |0|    ↓   |±∞|    ↓     |

(Again: you have to verify this table yourself!)

 

C. Local maximum is reached at x=0, f(0)=0

D.

Interval |(-∞, -2)| -2| (-2, 0)|0|(0, 2)| 2 |(2, +∞)|

  f''(x)    |    +     |na |    -     |0|    -   |na |    +     |

 

  f(x)      |     ∪   | ±∞|   ∩     |0|   ∩   |±∞|    ∪     |

Notations: ∪-concave, ∩-convex, na- indefinite,  ↑-increases, ↓-decreases.