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How do you evalulate $f^{-1}(5)$ where $f(x) = (3 + 2) - (x * 4)$

I understand that if $f(x) = y$ then $f^{-1}(y) = x$

The input and output are essentially reversed. The most common place I have seen and used this is in the trigonometric functions.

So if I wanted to evaluate $f^{-1}(5)$

My approach is to set $(3 + 2) - (x * 4) = 5$

Solve for $x = 0$

Therefore $f^{-1}(5) = 0$

But I was told by a math teacher a long time ago that you could evaluate the inverse of a function by plugging in the values and reversing the order of evaluation. For example, SADMEP instead of PEMDAS.

So I want to do any subtraction first on the expression: $(3 + 2) - (x * 4)$

I believe the first step is therefore $2 - 5$

Now I have $3 + (-3) * 4$

The next step is addition, and then multiplication (since there is no division).

$0 * 4$

$0$

The method apparently works, this time. Will this always work?

Also notice that I took the $2$ and subtracted $x$ or in this case $5$

I solved the expression from left to right, as opposed to doing it right to left. Is it also a requirement to still do things left to right? I believe left to right is still necessary, otherwise this would not have come out right (24 instead of 0).

I will probably make another question asking why $f^{-1}(x) = \frac{1}{f(x)}$ unless someone has a simple answer (you probably do).

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The last statement ($f^{-1}(x) = \frac{1}{f(x)}$) is false. $\dfrac{1}{\sin x} = \csc x \neq \sin^{-1} x$. The confusion may arise from assuming that $\sin^k x = (\sin x)^k$, but an exception is made for $k = -1$, where we define it as the inverse function. –  George V. Williams Jan 11 '13 at 2:45
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It is very rare that $f^{-1}(x) = \frac{1}{f(x)}$. You are thinking of the multiplicative inverse of a number $x$, which is $\frac{1}{x}$. When referring to functions, we want the compositional inverse of that function, which is significantly more complicated to express (in fact, it may not even exist). –  Austin Mohr Jan 11 '13 at 2:50
    
I see that it is false, I was very suspicious of that statement and now I see why I was mistaken. –  Leonardo Jan 11 '13 at 3:02
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1 Answer

up vote 1 down vote accepted

The general framework you describe (solve for $x$ in the equation $f(x) = y$) is correct. When you are solving for $x$, you are doing "reverse order of operations", which is what your teacher is calling SADMEP. It is a coincidence that you got the correct answer here (in fact, the first step of $3 + (-3)4$ is incorrect).

I'm not sure how to diagnose the problem except to show how I would do it and ask for your questions.

First, notice $$ (3 + 2) - (x \cdot 4) $$ is more concisely written as $$ 5 - 4x, $$ so we'll work with that. Now, we want to see what $x$ will give the output of $5$. That is, we want to solve $$ 5 - 4x = 5 $$ for $x$. Subtracting $5$ from both sides gives $$ -4x = 0, $$ and finally dividing both sides by $-4$ gives $$ x = 0, $$ as you've already found.

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So there is no reason to use SADMEP, and infact it is not a correct method? I think maybe if there is a correct SADMEP, it is very ambiguous and apparently beyond me at this point. That is, starting with subtraction and ending with exponents/parentheses is itself not correct. –  Leonardo Jan 11 '13 at 3:06
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@Leonardo "SADMEP" is just an acronym for "reverse order of operations". You seem to understand how to solve for $x$ in this way, but are complicating things by trying to force it directly onto the letters S A D M E P. Notice I subtracted $5$ before dividing by $-4$, as you did in your initial example. –  Austin Mohr Jan 11 '13 at 3:08
    
Yes I am taking it as an algorithmic method to evaluating, where it uses the reverse steps as PEMDAS. Thank you for clarifying why this SADMEP is not "literally" the opposite in the sense that I have considered it. –  Leonardo Jan 11 '13 at 3:10
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