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I have a function

$f(x) = (3*a)*x-3*b$

I know here the inverse function is

$f^{-1}(y) = (3*a)^{-1} * (y + 3b)$

I don't understand the steps that would lead to this or which math rules I may have forgotten.

wishi
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  • Hello. Could you provide more information on what you have tried? It's hard to help you without an intuition of where you may be having trouble. – lafinur Sep 28 '22 at 16:03
  • It's a while ago since I last did that. Tried to isolate the 3, checked for theorems (binomial etc.) . Pen and paper, not very useful I am afraid. – wishi Sep 28 '22 at 16:08
  • It is useful to have more specific information about why you are asking -- where you found the problem, why it is important to you -- details that presumably are well known to yourself but a matter of guesswork as you posed the problem to your Readers. – hardmath Sep 28 '22 at 16:29

1 Answers1

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Since I ignore your level in mathematics, I'll keep it very simple and not that rigorous.

The function $f$ takes an $x$ and returns a $y$. You want an inverse that takes a $y$ and returns and $x$ in a way that $x, y$ pairs are the same as defined by $f$. So simply call $f:=y$ and clear for $x$:

$$y = 3ax - 3b \implies y+3b=3ax \implies \frac{y+3b}{3a} = x$$

Now we want this to be a function, and we are not used to having functions defined in terms of $y$, so we change the names and say

$$\frac{x+3b}{3a} = y = f^{-1}(x)$$

More briefly,

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

which is what you have.

lafinur
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  • That gets me one step further in my recovery of forgotten knowledge. I don't understand why $\frac{x+3b}{3a}$ is equivalent to $\frac{1}{3a} (x+3b)$. I had an addition there instead of a multiplication. – wishi Sep 28 '22 at 16:33
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    $\frac{1}{3 \times a} = \frac{1}{3a}$ and $X \times \frac{1}{3a} = \frac{X}{3a}$, where $X$ is anything. In our case, $(x+3b) \times \frac{1}{3a} = \frac{(x+3b)}{3a}$, where the parenthesis are redundant and we simply have $ \frac{x+3b}{3a}$ – lafinur Sep 28 '22 at 16:53