Question

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

Answer 1

`h(x)=x+x^{ (1)/(2)} \\h=x+ √(x) \\x=t^2 \\h=t^2+t`
Solve for t


`t^2+t-h=0 \\ \\t= (-1\pm √(1+4h) )/(2) \\x=t^2=((-1\pm √(1+4h) )/(2) )^2`

The inverse function is


`h^(-1)(x)=((-1\pm √(1+4x) )/(2) )^2`

Now find
`h^(-1)(6)`


`h^(-1)(6)=((-1\pm √(1+4* 6) )/(2) )^2=((-1\pm 5 )/(2) )^2 \\ \\h^(-1)(6)=4\text{ or }h^(-1)(6)=9`