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What is the inverse of the function:

$$f(x)=\frac{x+2}{5x-1}$$

?

Answer:

$$f^{-1}(x)=\frac{x+2}{5x-1}$$

Can one of you explain how the inverse is the same exact thing as the original equation?

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  • $\begingroup$ not only this function but also $x, -x, 1/x, 1-x$ have the same property. $\endgroup$ – Bumblebee Aug 25 '14 at 9:46
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Let $f(x)=y$. So we have $y=\frac{x+2}{5x-1}$. Now "swap" the variables so we have $$x = \frac{y'+2}{5y'-1}$$ where $y' = f^{-1}(x)$ (to distinguish $y'$ from $y$). Now let's solve for $y'$. \begin{align} x &= \frac{y'+2}{5y'-1} \\ x(5y'-1) &= y'+2 & \text{multiply both sides by } 5y'-1 \\ 5xy' - x&=y'+2 & \text{distribute $x$ on the left side} \\ 5xy'-y'-x&=2 & \text{move all $y'$ to left side} \\ 5xy'-y'&=x+2 & \text{and add $x$ to right side} \\ (5x-1)y'&=x+2 & \text{factor out $y'$ on left side} \\ y'&=\frac{x+2}{5x-1} & \text{divide both sides by $5x-1$} \end{align} Thus, $f^{-1}(x)=y'=\frac{x+2}{5x-1}=y=f(x)$.

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  • $\begingroup$ understood-thankslot $\endgroup$ – MEE May 26 '14 at 20:17
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The inverse is not in general "the same exact thing as the original equation".

Generally, $f(x)\ne f^{-1}(x)$, but this is not always true. For example, consider the function $f(x) = -x$. This function is just the function that negates its input. Of course, if your negate your input twice, you get the original input. Put another way, to reverse the operation of negating your input, you simply negate it again.

It just happens to be the case that your function satisfies the same property. Namely, $$f(f(x)) = \frac{f(x)+2}{5f(x)-1} = \frac{\frac{x+2}{5x-1}+2}{5\frac{x+2}{5x-1} - 1} = x$$ and hence $f(x)$ is its own inverse. (ie, to reverse $f(x)$, simply apply it again!)

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    $\begingroup$ Most usefull answer. Especially for mentioning that $f(f(x))=x$ thus $f(x)=f^{-1}(x)$ instead of focus on the calculation. +1 $\endgroup$ – user127.0.0.1 May 26 '14 at 20:19
  • $\begingroup$ ok, i understand now-thanks $\endgroup$ – MEE May 26 '14 at 20:26
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    $\begingroup$ @MEE I want to point out that saying "English...please?" is very disrespectful to the person who is trying to help you understand. It implies that the reason you don't understand is the fault of the helper, that he is using a weird nonstandard language that nobody should be expected to understand. This is especially true on Q&A forums like this, where nobody is obligated in any way to help you. Also, you shouldn't expect to understand someone's answer in minutes. It's often the case in math that solutions aren't easy. Even elegant solutions may take hours/days to really understand. $\endgroup$ – oxeimon May 26 '14 at 21:01
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    $\begingroup$ sorry, sorry..totally understand, shouldn't have said-thanks for pointing that out for me :) And thank you @oxeimon $\endgroup$ – MEE May 26 '14 at 21:04
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    $\begingroup$ @MEE Saying "English...please?" also doesn't describe what you don't understand. If someone writes a couple paragraphs to explain something, and you don't understand, simply expressing that you don't understand is useless. Nobody is going to completely rewrite their carefully worded explanation to hopefully get something you will understand. What you should do is think about what exactly it is you don't understand, maybe a specific phrase or sentence, and ask about that. This is a key part of learning mathematics. $\endgroup$ – oxeimon May 26 '14 at 21:04
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Let $y=\frac{x+2}{5x-1}$. Then $5xy-y=x+2$, so $x(5y-1)=5xy-x=2+y$, so $x=\frac{2+y}{5y-1}$.

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First, note that this is not always the case.

As a geometric intuition, more than anything, consider the following: the inverse of a function, $f^{-1}(x)$, graphically, is a reflection of $y=f(x)$in the line $y=x$ (note that a function has an inverse iff it is injective). So, with regards to this statement, we should ask ourselves:"Under what circumstances is a function self-inverse (i.e. its own inverse)"?

The answer is: a function $f(x)=f^{-1}(x) \iff$ the graph $y=f(x)$ is symmetric about the line $y=x$, which is the case in this question, as we can see from this graph: enter image description here

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First we have $$f'(x)=-\frac{11}{(5x-1)^2}$$ hence $f$ is strictly decreasing on every interval $(-\infty,\frac15)$ and $(\frac15,+\infty)$ so its restriction has an inverse function on these intervals.

Now it suffices to verify that $f(f(x))=x$:

$$\frac{f(x)+2}{5f(x)-1}=\frac{\frac{x+2}{5x-1}+2}{5\frac{x+2}{5x-1}-1}=\frac{x+2+10x-2}{5x+10-5x+1}=x$$

and geometrically this means that the curve of $f$ is symmetric relative to the line with equation $y=x$.

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Write it as: $$(5x-1)y=x+2.\tag{$5x-1\neq0$}$$ Now replace all $x$s with $y\,$s: $$(5y-1)x=y+2.$$ Solving for $y$ gives rise to: $$y=\dfrac{x+2}{5x-1}.\tag{$5x-1\neq0$}$$ Which happens to be our inverse function.

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  • $\begingroup$ Asalaam Alykum and Thanks lot brother $\endgroup$ – MEE May 26 '14 at 20:23
  • $\begingroup$ @MEE Waalaykum Asalaam. Glad I could help! $\overset{\cdot\cdot}\smile$ $\endgroup$ – Hakim May 26 '14 at 20:24
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$$f(x) = \frac{x+2}{5x-1}$$ $$(5x-1)f(x)=x+2$$ $$5xf(x)-f(x)=x+2$$ $$5xf(x)-x=f(x)+2$$ $$x(5f(x)-1)=f(x)+2$$ $$x=\frac{f(x)+2}{5f(x)-1}$$

Functions which are there own inverses are called involutions.

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  • $\begingroup$ you-kind of lost me on the first step $\endgroup$ – MEE May 26 '14 at 20:20
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    $\begingroup$ @MEE Daniel just multiplied both sides by $(5x-1)$. $\endgroup$ – Hakim May 26 '14 at 20:23
  • $\begingroup$ got it thankssss $\endgroup$ – MEE May 26 '14 at 20:25

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