[tex]f(x)=4x^2\to y=4x^2\\\\\text{Interchange the y and the x}\\\\x=4y^2\\\\\text{Solve for y}\\\\4y^2=x\qquad|:4\\\\y^2=\dfrac{x}{4}\to y=\sqrt{\dfrac{x}{4}}\\\\y=\dfrac{\sqrt{x}}{\sqrt4}\\\\y=\dfrac{\sqrt{x}}{2}\\\\Answer:\ \boxed{f^{-1}(x)=\dfrac{\sqrt{x}}{2}}\ for\ x\geq0[/tex]
