If f(x) = 1/3x - 7, find f^-1(x).

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If `f(x)=1/(3x)-7`

Put `y=1/(3x)-7`

Interchange x and y,

`x=1/(3y)-7`

`rArr 1/(3y)=x+7`

`rArr 3y=1/(x+7)`

`rArr y=1/(3x+21)`

So, `f^(-1)(x)=1/(3x+21)`

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Check: `f(f^(-1)(x))=1/(3(1/(3x+21)))-7=1/(1/(x+7))-7=x+7-7=x`

`f^(-1)(f(x))=1/(3(1/(3x)-7)+21)=1/(1/x-21+21)=1/(1/x)=x`

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`f(x)=1/(3x)-7`

By def of inverse function

`f^(-1)(f(x))=x`

let

`y=f(x)=1/(3x)-7`

`therefore`

`f^(-1)(y)=x ` (i)

But

`y=1/(3x)-7`

`y+7=1/(3x)`

`3(y+7)=1/x`

`x=1/(3(y+7))`

sbstitute x in (i)

`f^(-1)(y)=1/(3(y+7))`

Thus inverse of f is

`f^(-1)(x)=1/(3(x+7))`

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