Hi!

My question is:

Given that f is a function defined by f(x) = (2x - 2) / (x^2 +x - 2)

a) For what values of x is f(x) discontinuous?
b) At each point of discontinuity found in part a, determine whether f(x) has a limit and, if so give the value of the limit.
c) write the equation for each vertical and each horizontal asymptote for f. Justify your answer.
d) A rational function g(x) = (a) / (b + x) is such that g(x)=f(x) whenever f is defined. Find the values of a and b.

Ok so I figured out the answers to a and b. A is "discontinuous at x = -2 and x = 1. and b is "as the limit approches -2, it does not exist and is nonremovable and as the limit approches 1, the limit is 2/3 and is removable". I'm not sure how to do part c and d though. Hopefully someone can help me!!

Thanks!! :)

I beg to differ with MathMate.

Since the limit exists at x=1, there is no vertical asymptote at x=1, only at x=-2

The statement, "the vertical asymptotes are found when the denominator becomes zero" should be clarified to say
"the vertical asymptotes are found when the denominator becomes zero but the numerator is non-zero"

You probably factored the function properly and got

f(x) = (2x - 2) / (x^2 +x - 2)
= 2(x-1)/[(x-1)(x+2)]
= 2/(x+2) , x not equal to 1

from f(x) = = 2(x-1)/[(x-1)(x+2)]
you are right to say that it is discontinuous at x = 1 and -2

notice when x = 1, f(1) = 0/0 which is indeterminate and
Limit = 2(x-1)/[(x-1)(x+2)] as x -->1
= 2/3
but when x=-2 , f(-2) = 2/0 which is undefined.

So we have a "hole" at (1,2/3) and an asymptote at x = -2

for the horizontal asymptote I let x --> ∞ in the original function
I see "large"/"really large" which goes to zero as x gets bigger.
so y = 0 is the horizontal asymptote.

for your last part, notice that our reduced function
f(x) = 2/(x+2) has the form a/(b+x), so
a = 2
b = 2

I will leave it up to you to fit all those parts in the proper question/answers.

(a) and (b) correct.

for (c), the vertical asymptotes are found when the denominator becomes zero. The vertical asymptotes are vertical lines passing through these singular values of x, in the form x=?.
The horizontal asymptote can be found (in this case) by finding the value of f(x) as x-> +∞ and x->-∞.
It will be in the form y=?.

(d) First look at f(x) as
f(x) = 2(x-1)/((x-1)(x+2))
When f(x) is defined, x≠1 and x≠-2. When x≠1, what can you say about the common factors (x-1) in the numerator and denominator?

Oh my goodness!!! Thank you so so much! That makes perfect sense! Thank you!!! :)

MathMate thank you so much as well!! The (x-1)'s cancel out which gives me the equation 2/(x+2), which like Reiny said, is in the form I need it in!

Seriously I cannot thank you enough for your help! I totally understand how to solve this problem now! :)

Reiny, thank you for the correction. That was an oversight.

To find the vertical and horizontal asymptotes of the function f(x) = (2x - 2) / (x^2 + x - 2), you need to examine the behavior of the function as x approaches positive or negative infinity.

c) Vertical Asymptotes:
Vertical asymptotes occur where the denominator becomes zero and the function becomes undefined. So, to find vertical asymptotes, set the denominator of f(x) equal to zero and solve for x.

x^2 + x - 2 = 0

This quadratic equation can be factored as:

(x + 2)(x - 1) = 0

From this, we can see that the function is undefined (discontinuous) at x = -2 and x = 1. Therefore, these are the vertical asymptotes of f(x).

d) Horizontal Asymptotes:
To determine the horizontal asymptote, we need to examine the behavior of the function as x approaches positive or negative infinity.

As x approaches positive or negative infinity, the highest power terms dominate the function. In this case, the highest power terms are x^2. As x approaches infinity, x^2 becomes much larger compared to other terms.

To determine the horizontal asymptote, we compare the degrees of the numerator and denominator polynomials.

In this function, the degree of the numerator is 1 (highest power of x) and the degree of the denominator is 2 (highest power of x^2). Therefore, the horizontal asymptote is a horizontal line with y = 0. The function f(x) approaches 0 as x approaches positive or negative infinity.

The vertical asymptotes are x = -2 and x = 1, and the horizontal asymptote is y = 0.

d) To find the values of a and b in the rational function g(x) = a / (b + x) such that g(x) = f(x) whenever f is defined, substitute the given function f(x) into g(x) and match the numerator and denominator.

From the given function f(x):

(2x - 2) / (x^2 + x - 2) = a / (b + x)

By comparing the numerators:

2x - 2 = a

By comparing the denominators:

x^2 + x - 2 = b + x

Rearranging this equation:

x^2 - x + b - 2 = 0

Comparing the coefficients, we get:

b - 2 = 0 (coefficient of x^0)
-1 = 0 (coefficient of x^1)

The second equation, -1 = 0, is not true, which means there is no value of b that satisfies this equation. Consequently, there are no values of a and b that make g(x) = f(x) for all x.