If `f:R->R` be defined by `f(x)=x^2+1`, then find `f^(-1)(17)` and `f^(-1)(-3)`.

Explanation

Let` f^(-1){17} = x `

`⇒ f(x) = 17 `

`⇒ x^2 + 1 = 17 `

`⇒ x^2 – 16 = 0 `

`⇒ (x – 4)(x + 4) = 0 ∴ x = +-4 `

Clearly, Both –4 and 4 are
elements of the domain R.
Thus, `f^(-1){17} = {–4, 4}
`

Now, Let `f^(-1){–3} = x`

...