Answer
$f^{-1}(x)=\dfrac{x^{2}-5}{8}$
Work Step by Step
$f(x)=\sqrt{5+8x}$
Rewrite this expression as $y=\sqrt{5+8x}$ and solve for $x$:
$y=\sqrt{5+8x}$
Square both sides:
$y^{2}=(\sqrt{5+8x})^{2}$
$y^{2}=5+8x$
Take $5$ to the left side:
$y^{2}-5=8x$
$8x=y^{2}-5$
Take the $8$ to divide the right side:
$x=\dfrac{y^{2}-5}{8}$
Interchange $x$ and $y$:
$y=\dfrac{x^{2}-5}{8}$
The inverse of the original function is $f^{-1}(x)=\dfrac{x^{2}-5}{8}$
