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If f(x)=3x-5, then f^(-1)(x) is given b...

If `f(x)=3x-5,` then `f^(-1)(x)` is given by `1/((3x-5))` is given by `((x+5))/3` does not exist because `f` is not one-one does not exist because `f` is not onto

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`y = f (x) = 3x – 5`
It is one-one onto function and hence it is invertible.
`y = 3x – 5`
`x = [y + 5] / 3`
...
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If f(x)=3x-5, then f^(-1)(x) is given by (a) (1)/((3x-5)) (b) ((x+5))/(3) (c)does not exist because f is not one-one (d)does not exist because f is not onto

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