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.
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!! :)
6 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?
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! :)
Seriously I cannot thank you enough for your help! I totally understand how to solve this problem now! :)
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"
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"
Reiny, thank you for the correction. That was an oversight.
1 answer