Find the inverse of each function. f(x) = √(2x - 3)

Solutions

VerifiedWrite the original function as an equation

$y=\sqrt{2x-3}$

To calculate inverse, switch $x$ and $y$ then solve for y.

$\begin{align*} x&=\sqrt{2y-3}\\ \\x^2&=(\sqrt{2y-3})^2&&\text{Squaring on both sides}\\ \\x^2&=2y-3\\ \\2y&=x^2+3\\ \\\dfrac{2y}{2}&=\dfrac{x^2+3}{2}\\ \\y&=\dfrac{1}{2}x^2+\dfrac{3}{2} \end{align*}$

The inverse function of $f(x)=\sqrt{2x-3}$ is $f^{-1}(x)=\dfrac{1}{2}x^2+\dfrac{3}{2}$

Write the original function as an equation by letting $f(x)=y$:

$y=\sqrt{2x-3}$

Switch $x$ and $y$ to switch domain and range:

$x=\sqrt{2y-3}$

Solve for $y$:

$x^2=2y-3$

$x^2+3=2y$

$\dfrac{1}{2}x^2+\dfrac{3}{2}=y$

The new $y$ is the inverse, $f^{-1}$. Because the range of $f$ is $y\geq 0$, then the domain of $f^{-1}$ must be restricted to $x\geq 0$ so we write:

$\color{#c34632}{f^{-1}(x)= \dfrac{1}{2}x^2+\dfrac{3}{2},x\geq 0}$

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