Where a Function Meets Its Inverse

The graph of $f^{-1}$ is the reflection of $f$ in the line $y = x$. So for an increasing function, wherever $f$ meets $f^{-1}$ it also meets $y = x$, and $f(x) = f^{-1}(x)$ reduces to the simpler $f(x) = x$.

The function $f$ is defined by $f(x) = 2x - 3$ for $x \in \mathbb{R}$. Find the value of $x$ for which $f(x) = f^{-1}(x)$. [4]

The working

Method 1, the shortcut (valid because $f$ is increasing): solve $f(x) = x$: $2x - 3 = x \implies x = 3 \quad \text{(M1, A1)}$

Method 2, directly. Find the inverse: let $y = 2x - 3$, swap and solve: $x = 2y - 3 \implies y = \frac{x + 3}{2}, \quad \text{so } f^{-1}(x) = \frac{x + 3}{2}. \quad \text{(M1)}$ Set $f(x) = f^{-1}(x)$: $2x - 3 = \frac{x + 3}{2} \implies 4x - 6 = x + 3 \implies 3x = 9 \implies x = 3 \quad \text{(A1)}$

Both methods give $x = 3$, the point $(3, 3)$ on $y = x$.

Where the marks are won and lost

• Knowing the reflection property ($f^{-1}$ is $f$ reflected in $y = x$) gives the fast route for increasing functions.
• If you find the inverse, do the algebra carefully: multiply both sides by $2$ before rearranging.
• The solution lies on $y = x$, so its coordinates are $(3, 3)$, a useful sense-check.

Common mistakes

• Applying $f(x) = x$ to a decreasing function, where meetings with $f^{-1}$ need not lie on $y = x$.
• Errors when forming the inverse (forgetting to swap $x$ and $y$).
• Arithmetic slips clearing the fraction $\frac{x+3}{2}$.

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