## Calculus 8th Edition

The inverse function is $$f^{-1}(x)=\frac{1}{2}+\frac{\sqrt{1+4x}}{2}.$$
Notice that $y=x^2-x=x(x-1)$ is a quadratic function. Its zeroes are $x_1=0$ and $x_2=1$. Further, we know that it will achieve its minimum in the point in the middle between its zeroes, i.e. when $x=1/2$ and the value of that minimum is $y_m=(1/2)^2-1/2=-1/4$. This means that in the given domain $x\geq1/2$ the function passed its minimum and is only increasing. Now we have to express $x$ in terms of $y$. $$y=x^2-x\Rightarrow x^2-x-y=0.$$ This is a quadratic equation whose solutions are given by $$x=\frac{1\pm\sqrt{1-4\cdot1(-y)}}{2}=\frac{1\pm\sqrt{1+4y}}{2}=\frac{1}{2}\pm\frac{\sqrt{1+4y}}{2}.$$ we have shown that $y\geq1/4$, so for $x$ to be greater than or equal to $1/2$ we must take the '+' sign only i.e. $$x=\frac{1}{2}+\frac{\sqrt{1+4y}}{2}.$$ This gives $$f^{-1}(y)=\frac{1}{2}+\frac{\sqrt{1+4y}}{2}.$$ Renaming $y$ back to $x$ this becomes $$f^{-1}(x)=\frac{1}{2}+\frac{\sqrt{1+4x}}{2}.$$