If `f: R->R` be defined by `f(x)=x^3-3` , then prove that `f^(-1)` exists and find a formula for `f^(-1)` . Hence, find `f^(-1)(24)` and `f^(-1)(5)` .

Explanation

We have f: `R arrow R` be defined by `f(x)=x^{3}-3`

So we prove that `{F}^{-1}` exists


Suppose x and y be 2 elements in domain (R),

Then ,` x^{3}-3=y^{3}-3`
`

Rightarrow {x}^{3}={y}^{3} `
`
Rightarrow {x}={y}`

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