Answer
$f^{-1}(x)=-\dfrac{1}{2}\pm\dfrac{\sqrt{4x+1}}{2}$
Work Step by Step
$f(x)=x^{2}+x$ $,$ $x\ge-\dfrac{1}{2}$
Rewrite this expression as $y=x^{2}+x$ and solve for $x$:
$y=x^{2}+x$
$x^{2}+x=y$
Take $y$ to the left side:
$x^{2}+x-y=0$
Solve for $x$ using the quadratic formula. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=1$ and $c=-y$.
Substitute the known values into the formula and simplify:
$x=\dfrac{-1\pm\sqrt{1^{2}-4(1)(-y)}}{2(1)}=\dfrac{-1\pm\sqrt{1+4y}}{2}=-\dfrac{1}{2}\pm\dfrac{\sqrt{4y+1}}{2}$
Interchange $x$ and $y$:
$y=-\dfrac{1}{2}\pm\dfrac{\sqrt{4x+1}}{2}$
The inverse of the original function is $f^{-1}(x)=-\dfrac{1}{2}\pm\dfrac{\sqrt{4x+1}}{2}$
