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Which of the following plots give the most accurate representation of the function $ f(x) = x/(x^2-1) $? These are the options

Attempt: at $ x=1\, f(x) = \infty $

at $ x= -1, f(x) = -\infty $

By that i arrive at the option d. Could anyone tell me if its the right answer?

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    $\begingroup$ What happens when we plug in numbers close, but less than $1$ like $.99$. What happens when we plug in numbers close, but more than $1$ like $1.1$? Option a) should be correct... $\endgroup$ Apr 6, 2017 at 4:09
  • $\begingroup$ The facts are right, the conclusion is wrong. The only graph where the limits at $\pm1$ are $\infty$ is a). On top it captures that $f(0)=0$ $\endgroup$
    – marwalix
    Apr 6, 2017 at 4:11
  • $\begingroup$ @AhmedS.Attaalla Thank You for pointing out what i failed to see, It helped me. $\endgroup$
    – rahul rj
    Apr 6, 2017 at 4:18

4 Answers 4

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The given function is: $f(x)=\frac {x}{x^2-1}$

Evaluating the function at x arbitrarily large, that is $\lim_{x\to\infty}f(x) = 0, \Rightarrow f(x)$ has a horizontal asymptote at $y=0$.

Furthermore, the vertical asymptote, (when values of $f(x) =y$ become arbitrarily large) is determined by considering where the given function is undefined.

Note $f(x) = \frac {x}{x^2-1}=\frac {x}{(x+1)(x-1)}\Rightarrow f(x)$ is undefined at $x = 1$ & $-1$. This thus implies when $x$ approaches $0$ from the left and right, its mapping $f(x)$ approaches infinity and negative infinity respectively.

To find any corresponding $x$-intercepts to the given function, we simply rationalize the function and solve for x on the numerator. Therefore, when $f(x) = 0, x=0$.

Thus finally we have two vertical asymptotes corresponding to the lines x=-1,1, a horizontal asymptote corresponding to the line y = 0, and a x-intercept at the point (0,0). Therefore the answer is figure a.

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Ok, so look at the denominator: $(x^2-1) = (x+1)(x-1)$. This means that there will be a vertical asymptote at $x=1$ and $x=-1$, as we cannot divide by zero. That eliminates all options except for a.

Notice that for d there is only one vertical asymptote, and thus this cannot be the answer.

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  • $\begingroup$ Aren't there two vertical asymptotes for option d?. Nevertheless I arrived at the option A putting a value less than 1 and greater than 1 as suggested in one of the comments. Thank you! $\endgroup$
    – rahul rj
    Apr 6, 2017 at 4:23
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    $\begingroup$ @rahulrj no, an asymptote is defined for a single point. In graph d, the graph approaches negative infinity from the left of x=0, and positive infinity from the right of x=0. But there's one asymptote: x=0. Hope that helps. $\endgroup$
    – user11088
    Apr 6, 2017 at 4:41
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    $\begingroup$ @Hamlet Indeed that explanation was immensely helpful. It seemed I had a wrong idea of what an asymptote was. $\endgroup$
    – rahul rj
    Apr 6, 2017 at 4:49
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$f(x) = \frac{x}{(x-1)(x+1)}$. When x equals one or negative one, the denominator equals zero. You can't divide by zero, so there f can not have a value when x is positive or negative number.

If you look at the different options, you'll see that only one graph has this property. Graph number a does not have values at positive or negative one; when x approaches -1 or 1 the graph shoots towards positive and negative infinity (depending on the direction you approach -1 or 1).

Graph d is an incorrect answer. It's easy to see this: $f(0)=0$, but that isn't true for graph d.

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I approached it in a different way, so I wanted to add this for completeness. It does use a little calculus.

It should be easy to see (in fact, immediately obvious) that the function given is the first derivative of $\frac 12 \ln |x^2-1|$. The constant coefficient is not very important since we're looking at trends and asymptotes.

First we need to sketch $\ln|x^2-1|$. Sketching $|x^2-1|$ is trivially easy, after which the basic properties of the logarithm function can be applied to the curve, e.g. $\ln 1 = 0, \lim_{x \to 0} \ln x = -\infty$, and so forth.

The final step is just observing how the gradient changes and sketch that.

It sounds involved, but it's not that complicated when you actually do it.

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