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Why does the function: $$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$$

Is defined at $x = -1, x = 0$ (we will have $1/0$ in the fraction then) and not at $x = 1$ according to the following graph?

enter image description here

Live example

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  • $\begingroup$ If you substitute 1 in $f(x)$ then it does not exits as $\frac{1}{0}$ does not exist $\endgroup$
    – tien lee
    May 2, 2018 at 14:38
  • $\begingroup$ @tienlee yep, but why is it defined at -1 and 0 then? $\endgroup$
    – techkuz
    May 2, 2018 at 14:39
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    $\begingroup$ It is not! If it is then tell me how much is $f(0)$, without graph $\endgroup$
    – nonuser
    May 2, 2018 at 14:39
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    $\begingroup$ @trthhrtz This function is not defined at $-1$ and $0$, but it does have removable discontinuities at $-1$ and $0$, while the discontinuity at $1$ is not removable. Are you familiar with the concept of removable discontinuities? $\endgroup$
    – BallBoy
    May 2, 2018 at 14:41
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    $\begingroup$ If you algebraically manipulate your function, you'll find it is 'equivalent' (note the quotations) to $$f(x) = \frac{2x^{2}}{(3x+1)(x-1)}$$ which is what your graph is of. $\endgroup$ May 2, 2018 at 14:42

5 Answers 5

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Technically, it is not defined. When you graph the function on a computer, it doesn't show it, but there should really be a "hole" at $-1$ and $0$ because we are indeed trying to evaluate something that contains $\frac 10$ (something we call a "singularity").

However, the reason it seems to be defined at $-1$ and $0$ is because these are classified as "removable singularities", which means that although there is a singularity at that point, the limit of the function as $x$ tends to that point of that point still exists.

In this case:

\begin{align} \ \lim_{x \rightarrow 0}\frac{\frac{1}{x-1}+\frac{1}{x+1}}{\frac{2}{x+1}+\frac{1}{x}} & = \lim_{x \rightarrow 0}\frac{\frac{x}{x-1}+\frac{x}{x+1}}{\frac{2x}{x+1}+1}=\frac 01 =0 \end{align}

Similarly with $\lim_{x\rightarrow -1}$.

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Let's simplify your fraction a little bit. Multiply the top and bottom by $(x-1)x(x+1)$ to get

$$\dfrac{x(x+1) + (x-1)x}{2(x-1)x+(x-1)(x+1)} = \dfrac{2x^2}{3x^2-2x-1} = \dfrac{2x^2}{(3x+1)(x-1)}$$

This new function is the same as your old function everywhere except at $-1,0,1$ (where we can't multiply the top and bottom by $(x-1)x(x+1)$, since then we'd be multiplying by $\frac00$), and it's defined everywhere except at $1$ and $-\frac13$. In particular it's defined at $-1$ and $0$.

Your original function isn't defined at $-1$ and $0$, but it does have "removable discontinuities" there. I'm not sure how much calculus you know, but loosely, this means that to get from the new function $\dfrac{2x^2}{(3x+1)(x-1)}$ to your original function, you just remove the single points at $-1$ and $0$. This doesn't cause any asymptotes like you get at $-\frac13$ and $1$.

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$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$

$f(x)=\frac{(x-1)(x+1)(x)}{(x-1)(x+1)(x)} \cdot \frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}} $

$f(x)=\frac{(x+1)(x)+(x-1)(x)}{2(x-1)(x)+(x-1)(x+1)}$

$f(x)=\frac{x^2+x+x^2-x}{2x^2-2x+x^2-1}$

$f(x)=\frac{2x^2}{3x^2-2x-1}$

$f(x)=\frac{2x^2}{(3x+1)(x-1)}$

So the vertical asymptotes at $x=-\frac{1}{3}$ and $x=1$ while there only holes at $x=0$ and $x=-1$

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$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$

$=\frac{\frac{2x}{(x-1)(x+1)}}{\frac{3x+1}{x(x+1)}}$

$=\frac{2x^2(x+1)}{(x-1)(x+1)(3x+1)}$

$=\frac{2x^2}{(x-1)(3x+1)}$

so $f(x)$ can be defined except at $x=1$ and $x=-\frac{1}{3}$. The apparent singularities at $x=0$ and $x=-1$ are removable.

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Let consider

$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$$

from here we need to set $x\neq1,-1,0$, morover we need that

$$\frac{2}{x+1} + \frac{1}{x}=\frac{3x+1}{x(x+1)}\neq 0\implies x\neq -\frac13$$

Simplifing the original expression we obtain

$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}\to\bar f(x)=\frac{2x^2}{(3x+1)(x-1)}$$

which is well defined for $x\neq -\frac13,1$.

In conclusion the function $f(x)$ is not defined for $x=1,-1,0,-\frac13$ and $x=0$ and $x=-1$ are two point with removable discontinuity.

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