# Inverse of x/(xˆ2-4)

• Jul 28th 2011, 04:15 PM
nicksbyman
Inverse of x/(xˆ2-4)
Attempted solution:

y=x/(xˆ2-4)

let y = inverse of y

x=y/(yˆ2-4)

xyˆ2-4x = y

xyˆ2-y-4x = 0

a = x, b = -1, c = -4x

y = (1 +/- sqrt(1-(4x)(-4x)))/(2x)

= (1+/- sqrt(1+16xˆ2))/2x, which is my solution

actual solution is:

y= (1-sqrt(16+xˆ2))/2x when x≠0
y= 0 when x = 0

I don't understand why the + sign has been discarded from the solution when x≠0 and I don't understand why y = 0 when x = 0. You could set y to 0 and solve for x in the original equation, but is that what is supposed to be done?

Thanks.
• Jul 28th 2011, 04:33 PM
Plato
Re: Inverse of x/(xˆ2-4)
Quote:

Originally Posted by nicksbyman
Attempted solution:
y=x/(xˆ2-4)

What in the world would lead you to think that that has an inverse?
It does not have an inverse!

Graph the function and look at it.

It has an inverse on $\displaystyle (-\infty,-2),\text{ or }(-2,2),\text{ or }(2,\infty).$
• Jul 28th 2011, 04:43 PM
mathjeet
Re: Inverse of x/(xˆ2-4)
Yes, you need to take a look at domain and range of function and if it is one to one then it has inverse.
• Jul 28th 2011, 04:48 PM
TKHunny
Re: Inverse of x/(xˆ2-4)
You simply must consider things up front so that they are not a surprise along the way.

Start from the top:

y = x/(x^2 - 4) perhaps it is wiser to think about f(x) = x/(x^2 - 4). This may help as we think on Domain and Range.

1) Discard x = 2 and x = -2. The Domain of f(x) is the Real Numbers with 2 and -2 excluded.
2) y = 0 is an asymptote for the graph. One might be tempted di discard y = 0 from the Domain, but remember that asymptotes don't really care about more centralized behavior. We have f(0) = 0.
3) Having determined that the function f(x) is in three pieces, one is caused to wonder if there is an easy INVERSE FUNCTION. A quick check of the graph indicates there is not. We'll have to chop it up a bit. Vertical asumptotes are a great place to start making divisions of the Range. There may be entirely different solutions for inverses of the three pieces. It's possible that two pieces will work together reasonably well. Thinking on it a bit, I think the middle portion should have one solution and the two outer portions another.
4) Very nicely, you have found the inverse relation of the two ourter portions. Good work. The Domain of the outer portions is x < -2 or x > 2. Notice how [-2,2] is not included. The range of the outer portions is all Real Numbers except y = 0. we can expect, then, that a proper inverse would have a Domain of all Real Numbers with zero (0) removed and a Range of y>2 or y < -2. See if these last characteristics fit your solution with the "+". Notice how, since x is already not zero, it is nto necessary to exclude it artificially and it is inappropriate to define it.
5) You "Actual Solution" seems to be missing some stuff. It has the inverse of only the middle section of f(x). If this is what is wanted, then the "actual solution" is right on.

Really, you have to think about these things. You can't just swap and solve willy-nilly.
• Jul 28th 2011, 04:50 PM
Plato
Re: Inverse of x/(xˆ2-4)
Quote:

Originally Posted by mathjeet
Yes, you need to take a look at domain and range of function and if it is one to one then it has inverse.

Oh my goodness, come on:
$\displaystyle f(x)=\frac{x}{x^2-4}$ has no inverse!
• Jul 28th 2011, 04:55 PM
mathjeet
Re: Inverse of x/(xˆ2-4)
Sorry
I didn't say it has no inverse but said that check the graph for horizontal and vertical line test and if it passes the test then it has an inverse.
• Jul 28th 2011, 05:02 PM
nicksbyman
Re: Inverse of x/(xˆ2-4)
I'm terribly sorry all I can't believe I missed this. The question asks for the solution on the open interval (-2, 2)
• Jul 28th 2011, 08:30 PM
mr fantastic
Re: Inverse of x/(xˆ2-4)
Quote:

Originally Posted by nicksbyman
I'm terribly sorry all I can't believe I missed this. The question asks for the solution on the open interval (-2, 2)

$\displaystyle f^{-1}(x)= \frac{1 \pm \sqrt{16x^2+1}}{2x}$ (assuming what you wrote has a typo, by the way).

Clearly (1, -1/3) is a point on the graph of y = f(x). Therefore (-1/3, 1) is a point on the graph of $\displaystyle y = f^{-1}(x)$. So the positive root in the above is rejected. Therefore

$\displaystyle f^{-1}(x)= \frac{1 - \sqrt{16x^2 + 1}}{2x}$

Clearly (0, 0) is a point on the graph of y = f(x). Therefore (0, 0) is a point on the graph of $\displaystyle y = f^{-1}(x)$. But $\displaystyle f^{-1}(0)$ has the indeterminant form 0/0. Therefore you must define what happens at x = 0 separately.
• Jul 28th 2011, 10:52 PM
TKHunny
Re: Inverse of x/(xˆ2-4)
Quote:

Originally Posted by nicksbyman
I'm terribly sorry all I can't believe I missed this. The question asks for the solution on the open interval (-2, 2)

Indeed. See #5 :-)
• Jul 29th 2011, 05:37 AM
chisigma
Re: Inverse of x/(xˆ2-4)
Quote:

Originally Posted by mr fantastic

$\displaystyle f^{-1}(x)= \frac{1 \pm \sqrt{16x^2+1}}{2x}$ (assuming what you wrote has a typo, by the way).

Clearly (1, -1/3) is a point on the graph of y = f(x). Therefore (-1/3, 1) is a point on the graph of $\displaystyle y = f^{-1}(x)$. So the positive root in the above is rejected. Therefore

$\displaystyle f^{-1}(x)= \frac{1 - \sqrt{16x^2 + 1}}{2x}$

Clearly (0, 0) is a point on the graph of y = f(x). Therefore (0, 0) is a point on the graph of $\displaystyle y = f^{-1}(x)$. But $\displaystyle f^{-1}(0)$ has the indeterminant form 0/0. Therefore you must define what happens at x = 0 separately.

As explained in...

http://www.mathhelpforum.com/math-he...ns-185019.html

... if exists the limit...

$\displaystyle \lim_{x \rightarrow 0} \frac{x}{f(x)}= a_{1}\ne 0$ (1)

... then $\displaystyle f(*)$ has inverse function $\displaystyle f^{-1}(*)$ and is...

$\displaystyle \lim_{x \rightarrow 0} \frac{f^{-1}(x)}{x}= a_{1}$ (2)

In our case is $\displaystyle f(x)= \frac{x}{x^{2}-4}$ so that the limit (1) is $\displaystyle a_{1}=-4$. Now is...

$\displaystyle \lim_{x \rightarrow 0} \frac{1-\sqrt{16\ x^{2}+1}}{2\ x^{2}}= \lim_{x \rightarrow 0} \frac{-16\ x^{2}}{2\ x^{2}\ (1+\sqrt{16\ x^{2}+1})} = -4$ (3)

... so that the result is 'all right' ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$