Question

If f(x)=(x-8)^2. for x≥8, find the inverse of f, f^-1(x).

Answer 1

If
`f^(-1)(x)` is the inverse of
`f(x)=(x-8)^2` for
`x\ge8`, then for
`f^(-1)(x)\ge8` we have

`f\left(f^(-1)(x)\right) = \left(f^(-1)(x)-8\right)^2 = x \\\\ \sqrt{\left(f^(-1)(x)-8\right)^2} = √(x) \\\\ f^(-1)(x)-8 = √(x) \\\\ \boxed{f^(-1)(x) = 8+√(x)}`

More precisely,

`\sqrt{\left(f^(-1)(x)-8\right)^2} = \left|f^(-1)(x)-8\right|`

but since the inverse is at least 8, the quantity inside the absolute is non-negative, so

`\sqrt{\left(f^(-1)(x)-8\right)^2} = f^(-1)(x)-8`