Question

Need help: If h(x) = x + x^(1/2), then find h^(-1)(6).

Answer 1

$h(x)=x+x^{ \frac{1}{2}} \\h=x+ \sqrt{x} \\x=t^2 \\h=t^2+t$
Solve for t

$t^2+t-h=0 \\ \\t= \frac{-1\pm \sqrt{1+4h} }{2} \\x=t^2=(\frac{-1\pm \sqrt{1+4h} }{2} )^2$

The inverse function is 

$h^{-1}(x)=(\frac{-1\pm \sqrt{1+4x} }{2} )^2$

Now find $h^{-1}(6)$

$h^{-1}(6)=(\frac{-1\pm \sqrt{1+4\times 6} }{2} )^2=(\frac{-1\pm 5 }{2} )^2 \\ \\h^{-1}(6)=4\text{ or }h^{-1}(6)=9$