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?
On
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.
On
$f(x) = \frac{x}{(x-1)(x+1)}$. 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.
On
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.
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.