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Prove the following.
If the function f(x) = √(2x - 3) is invertible then find its inverse. Hence prove that (f ∘ f⁻¹)(x)…
If the function $f(x) = \sqrt{2x - 3}$ is invertible then find its inverse. Hence prove that $(f \circ f^{-1})(x) = x$.
To show: $(f \circ f^{-1})(x) = x$
Answer
Answer
$f^{-1}(x) = \frac{x^2+3}{2}$, $x \ge 0$.
Hence $(f\circ f^{-1})(x) = x$.
- Let $y = f(x) = \sqrt{2x-3}$, so $y \ge 0$, by the definition of $f$
- Squaring: $y^2 = 2x - 3$, so $x = \frac{y^2+3}{2}$
- Interchanging $x$ and $y$: $f^{-1}(x) = \frac{x^2+3}{2}$, defined for $x \ge 0$
- $(f\circ f^{-1})(x) = f\left(\frac{x^2+3}{2}\right)$, by the definition of composition
- $= \sqrt{2\cdot\frac{x^2+3}{2} - 3}$, substituting in $f$
- $= \sqrt{x^2+3-3} = \sqrt{x^2}$, simplifying
- $= x$, since $x \ge 0$
- Hence $(f \circ f^{-1})(x) = x$