Answer
$f^{-1}(x) = \frac{1}{4}x^{2} + \frac{7}{4}$
Work Step by Step
$$f(x) = \sqrt {4x - 7}$$
To find $f^{-1}(x)$, we substitute $f(x)$ for an arbitrary variable $y$, and switch the positions of $x$ and $y$ like so:
$y = \sqrt {4x - 7}$
$x = \sqrt {4y - 7}$
Next, we solve for y:
$x^{2} = (\sqrt {4y - 7})^{2}$
$x^{2} = 4y - 7$
$x^{2} + 7 = 4y$
$\frac{x^{2} + 7}{4} = \frac{4y}{4}$
$\frac{1}{4}x^{2} + \frac{7}{4} = y$
Finally, $y$ becomes $f^{-1}(x)$:
$f^{-1}(x) = \frac{1}{4}x^{2} + \frac{7}{4}$
