Note: `f@g(x) = f(g(x))`

This means, where you used to see an "x" in the equation for f(x), now plug in "g(x)"

To find `f^(-1)(x)`

First start with your equation, then switch the x and the y, then solve for y.

So for example:

`f(x)=4x+3`

Think of this as: `y=4x+3`

To find the...

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Note: `f@g(x) = f(g(x))`

This means, where you used to see an "x" in the equation for f(x), now plug in "g(x)"

To find `f^(-1)(x)`

First start with your equation, then switch the x and the y, then solve for y.

So for example:

`f(x)=4x+3`

Think of this as: `y=4x+3`

To find the inverse, switch the x and the y:

`x=4y+3`

Then solve for y

`y=(x-3)/4`

With this in mind, start with the first pair of functions:

`f(x)=x-2`

`g(x)=x^2`

`f@g(x)=x^2-2`

`(f@g)^(-1)(x)=sqrt(x+2)`

`f^(-1)(x)=x+2`

`g^(-1)(x)=root()(x)`

`f^(-1) @ g^(-1) (x) = root()(x)+2`

`g^(-1)@f^(-1)(x)=root()(x+2)`

So, at least in this case,

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

Do the same for the second set of equations:

`f(x)=(1)/(x)+1`

`g(x)=3x-2`

`f@g(x)=(1)/(3x-2)+1`

`(f@g)^(-1)(x)=``((1)/(x-1)+2)/(3)=(2x-1)/(3(x-1)`

` `

` `

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

`g^(-1)(x)=(x+2)/(3)`

`f^(-1)@g^(-1)(x)=(1)/((x+2)/(3)-1)=(1)/((x-1)/(3))=(3)/(x-1)`

`g^(-1)@f^(-1)(x)=((1)/(x-1)+2)/(3)=((2x-1)/(x-1))/3=(2x-1)/(3(x-1)`

` `

So again, we have:

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

PS: We have only shown

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

is true for these pairs of functions, but it is always true