Question

Please help?

Answer 1

Answer c). See the explanation below.

Explanation:

We have `f(x)=x³+5`
So f: `RR->RR`
Now let `y in RR`:
`y=x³+5`
Let's find `x=f^-1(y)`.
`y=x³+5`
`y-5=x³`
`x=^3sqrt(y-5)`
And `f^-1(y):RR->RR`
\0/ here's our answer!