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**Short Answer**  
Yes,  
\[
f^{-1}(x) \;=\; -\tfrac12\,x \;+\;\tfrac32
\]  
really *is* the inverse of \(f(x)=-2x+3\), and having a fractional intercept is perfectly normal.

---

## Why the Fractional Intercept Is Correct

### 1. The Usual Steps for Finding the Inverse

1. **Start** with the function:  
   \[
   f(x) \;=\; -2x \;+\; 3.
   \]  
2. **Rename** \(f(x)\) to \(y\):  
   \[
   y \;=\; -2x \;+\; 3.
   \]  
3. **Swap** \(x\) and \(y\):  
   \[
   x \;=\; -2y \;+\; 3.
   \]  
4. **Solve** for \(y\):  
   \[
   -2y \;=\; x - 3
   \quad\Longrightarrow\quad
   y \;=\; \frac{x - 3}{-2} \;=\; -\tfrac12\,x \;+\;\tfrac32.
   \]

Hence,  
\[
\boxed{f^{-1}(x) = -\tfrac12\,x + \tfrac32.}
\]

### 2. Reflection Across \(y=x\)

Algebraically, an inverse function “undoes” what the original does.  Geometrically, the graph of an inverse function is the reflection of the original graph across the line \(y = x\).

- The **original** line has a slope of \(-2\) and a \(y\)-intercept of \(3\).  
- Its **inverse** has a slope of \(-\tfrac12\), which is indeed the “negative reciprocal” reflection across \(y=x\).  And the new intercept is \(\tfrac32\).

Those seemingly “odd” fractions are exactly what you get when you reflect across \(y=x\); there’s **nothing wrong** with that—lines can certainly have fractional intercepts!

### 3. Check by Composition

To be absolutely sure it’s correct, plug \(f^{-1}(x)\) *into* \(f\) and verify it simplifies to \(x\):

\[
\begin{aligned}
f\bigl(f^{-1}(x)\bigr)
&=\;
-2\Bigl(-\tfrac12\,x + \tfrac32\Bigr) + 3 \\
&=\;
-2\Bigl(-\tfrac12\,x\Bigr)
\;-\;2\Bigl(\tfrac32\Bigr)
\;+\;3
\\
&=\;
\bigl(+ x \bigr)
\;-\;3
\;+\;3
\\
&=\;
x.
\end{aligned}
\]
This confirms that \(\,f^{-1}(x)=-\tfrac12\,x + \tfrac32\) *is* the correct inverse function.

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### Final Takeaway

Having a fractional intercept is not a sign of any mistake; it’s exactly what you expect when a line of slope \(-2\) is reflected across \(y=x\). The inverse does *not* need to share the same integer intercept; rather, it should “flip” the roles of \(x\) and \(y\), which is precisely what you found.

Sat, 11 Jan 2025 03:52 GMT