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I'm having an issue with this problem for solving for inverse function:

$$f(x) = \frac{9x + 5}{x + 4}$$

Step 1: f(x) to Y. Then, change "$x$" to "$y$" in all cases.

$$f(x) = \frac{9y + 5}{y + 4}$$

Step 2. Multiply denominator to the other side whereas $x$ is $x(y + 4)$, distribute, $xy + 4x$.

Step 3: (This is where I think I am wrong). Divide left side by $x$ to get, $\frac{y + 4}{x}$.

$$\frac {y + 4}{x} = 9y + 5$$

I'm at a loss here, I've looked on youtube and other websites but none of them really explain a problem like this. I know to solve for "$y$", but I can't seem to isolate it.

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3 Answers 3

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In order to find the inverse of the function

$$f(x) = \frac{9x + 5}{x + 4}$$

First, replace $f(x)$ with $y$ to get:

$$y = \frac{9x + 5}{x + 4}$$

Then, switch the $y$ for the $x$'s and vice versa.

$$x = \frac{9y + 5}{y + 4}$$

Now, solve for $y$.

\begin{align} x (y + 4) &= 9y + 5 \\ xy + 4x &= 9y + 5 \\ 4x - 5 &= 9y - xy \\ 4x - 5 &= y(9 - x) \\ \frac{4x - 5}{9 - x} &= y \end{align}

The mistake you're making is dividing when you should just be subtracting the $xy$ once you have distributed it out.

Remember, you are solving for $y$ which means you have to bring your $y$'s all to one side as you would do when you are solving for $x$, for instance. Let me know if you need me to clarify further.

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    $\begingroup$ Wouldn't you subtract 5 from the right side instead of add? 4x - 5 / 9 -x ? $\endgroup$
    – mydiax
    Feb 26, 2016 at 2:50
  • $\begingroup$ Yes, good catch! My mistake :P I thought something was fishy. $\endgroup$
    – Jeel Shah
    Feb 26, 2016 at 2:51
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I don't know what you are doing with the replace $x$ by $y$ step.

You have $y = {9x +5 \over x+4 }$ or $yx+4 y = 9x + 5$, which gives $xy -9x = x(y-9)= 5-4y$, so we have $x = {5-4y \over y-9 }$. So the inverse function is $f^{-1} (y)= {5-4y \over y-9 }$.

Note that in general, it is not straightforward to compute the inverse function.

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If you divide $xy+4x$ by $x$ you do not get $\frac{y+4}x$.

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