Finding the inverse of a linear function from its equation

Key takeaway: This problem focuses on recognizing a one-to-one linear function and rewriting it in inverse form by exchanging the roles of x and y.

What this function is asking you to do

The given function is a linear function, $y=3-x$. To find its inverse, you swap the inputs and outputs and then solve for the new output variable.

Because this is a linear function with a nonzero slope, it is one-to-one and has an inverse. The inverse will also be linear.

Step-by-step inverse process

Start with the original equation:

$y=3-x$

Now swap $x$ and $y$:

$x=3-y$

Solve for $y$ by subtracting 3 from both sides:

$x-3=-y$

Multiply both sides by $-1$:

$y=3-x$

So the inverse function is

$f^{-1}(x)=3-x$

Key property to notice

This function is its own inverse. That happens because the rule $f(x)=3-x$ reflects every input across the line $y=x$ and lands back on the same algebraic rule.

A quick check confirms it:

$f(f(x))=3-(3-x)=x$

So applying the function twice returns the original value.

Common mistake to avoid

A frequent mistake is to stop after swapping $x$ and $y$ and write the inverse as $x=3-y$ instead of solving for $y$. The inverse must be written as a function of $x$, not as an equation with both variables still mixed together.

Another mistake is to think every inverse must look different from the original. Some functions, including this one, are self-inverse. That is not an error; it is a real algebraic property.

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Pitfalls the pros know 👇 One common pitfall is forgetting that an inverse must be written in function form. After swapping $x$ and $y$, you still need to isolate the new output variable. If you leave the result as $x=3-y$, you have not finished the inverse. Another issue is confusing inverse notation with a reciprocal. The inverse function of $f(x)=3-x$ is not $1/(3-x)$; it is the rule that undoes the original mapping. Always verify by composition if the result seems surprising.

What if the problem changes? If the problem were changed to $f(x)=5-x$, the same method would apply: swap $x$ and $y$ and solve for $y$. You would get $x=5-y$, then $y=5-x$, so the inverse would still be $f^{-1}(x)=5-x$. If the variant were $f(x)=3x-6$, then the inverse would no longer match the original; you would need to solve $x=3y-6$ for $y$ and get $f^{-1}(x)=\frac{x+6}{3}$. That shows how the inverse changes when the slope is not $-1$.

Tags: one-to-one function, inverse function, composition of functions