Find a formula for the inverse function?

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

2 Answers
Nov 12, 2017

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)

Nov 12, 2017

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)