Question

Find a formula for the inverse function?

I think that you multiply a number on each side but I'm really confused on how to find the inverse?

Answer 1

See explanation.

Explanation:

If you want to find an inverse function of

`y=(3x+2)/(7x-6)`

first you can multiply both sides by the denominator:

`y(7x-6)=3x+2` ##

`7xy-6y=3x+2`

Now we can rearrange the equation so that the new unknown `x` is on the left side:

`7xy-3x=6y+2`

`x*(7y-3)=6y+2`

`x=(6y+2)/(7y+3)`

Now we can change the variables back. This causes that `x` becomes the argument and `y` the function value (the usual naming convention)

The inverse function is:

`y=(6x+2)/(7x+3)`

Answer 2

Substitute `h^-1(x)` for every instance of `x` within `h(x)`, this causes the left side to become x, then solve the equation for `h^-1(x)`.

Explanation:

Given: `h(x) = (3x+2)/(7x-6)`

Substitute `h^-1(x)` for every instance of `x`:

`h(h^-1(x)) = (3(h^-1(x))+2)/(7(h^-1(x))-6)`

One of the two parts of the definition of an inverse is that `h(h^-1(x)) = x`, therefore, the left side becomes x:

`x = (3(h^-1(x))+2)/(7(h^-1(x))-6)`

Multiply both sides by `(7(h^-1(x))-6)`:

`x(7(h^-1(x))-6) = 3(h^-1(x))+2`

Distribute the x:

`7x(h^-1(x))-6x = 3(h^-1(x))+2`

Add 6x to both sides:

`7x(h^-1(x)) = 3(h^-1(x)) + 6x+2`

Subtract `3(h^-1(x))` from both sides:

`7x(h^-1(x)) - 3(h^-1(x)) = 6x+2`

Factor out `h^-1(x)`:

`(7x - 3)h^-1(x) = 6x+2`

Divide both sides by `(7x - 3)`:

`h^-1(x) = (6x+2)/(7x - 3)`

To verify that this is truly the inverse, we must test that `h(h^-1(x)) = x` and `h^-1(h(x)) = x`.

Start verification with `h(h^-1(x))`:

`h(h^-1(x)) = (3((6x+2)/(7x - 3))+2)/(7((6x+2)/(7x - 3))-6)`

Multiply the right side by 1 in the form of `(7x - 3)/(7x - 3)`:

`h(h^-1(x)) = (7x - 3)/(7x - 3)(3((6x+2)/(7x - 3))+2)/(7((6x+2)/(7x - 3))-6)`

This has the effect of remove the embedded fractions and distributing the denominator over the constants:

`h(h^-1(x)) = (3(6x+2)+2(7x - 3))/(7(6x+2)-6(7x - 3))`

Use the distributive property to perform 4 multiplications:

`h(h^-1(x)) = (18x+6+14x - 6)/(42x+14-42x + 18)`

Combine like terms:

`h(h^-1(x)) = (32x)/(32)`

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

Finish the verification with `h^-1(h(x))`:

`h^-1(h(x))= (6((3x+2)/(7x-6))+2)/(7((3x+2)/(7x-6)) - 3)`

Multiply by 1 in the form of `(7x-6)/(7x-6)`:

`h^-1(h(x))= (7x-6)/(7x-6)(6((3x+2)/(7x-6))+2)/(7((3x+2)/(7x-6)) - 3)`

Eliminate the embedded fractions and distribute the numerator among the constants:

`h^-1(h(x))= (6(3x+2)+2(7x-6))/(7(3x+2) - 3(7x-6))`

Use the distributive property to perform 4 multiplications:

`h^-1(h(x))= (18x+12+14x-12)/(21x+14 - 21x+18)`

Combine like terms:

`h^-1(h(x))= (32x)/(32)`

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

It is verified that `h^-1(x) = (6x+2)/(7x - 3)`