f(x) = (3x+5)/(3x-5)

To find the inversem, let f(x) = y

y = (3x+5)/(3x-5)

Cross multiply:

y(3x-5) = 3x + 5

3xy - 5y = 3x + 5

3xy - 3x = 5y + 5

factor out x:

x (3y - 3) = 5y + 5

divide by 3y-3

==> x = (5y+5)/(3y-3)

Then the inverse is:

==> f(x)^-1 = (5x+5)/(3x-3)

We'll re-write the function in a convenient way:

f(x)=(3x+5)/(3x-5) as y=(3x+5)/(3x-5)

Now, we'll solve this equation for x, multiplying both sides by (3x-5):

3xy-5y = (3x+5)

We'll move all terms containing x, to the left side and all terms in y, to the right side:

3xy-3x = 5+5y

We'll factorize:

x(3y-3) = 5(1+y)

x= 5(1+y) / (3y-3)

Now, we'll interchange x and y:

y = 5(1+x) / (3x-3)

So, the inverse function is:

**[f(x)]^(-1) = 5(1+x) / (3x-3)**

f(x) = (3x+5)/(3x-5). To find the inverse.

We shall find the inverse by solving for x and we see that f(x) o doesnot exist for 3x = 5.

f(x) = y =( 3x+5)/(3x-5)

Multiply by 3x-5. And let us try to make x subject.

y(3x-5) = 3x+5

Collect x's together.

y*3x-3x = 5+5y

3x(y-1) =5(y+1)

x = 5(y+1)/{3(y-1)} . Swapping, x and y,

y = 5(x+1)/{3(x-1)} is the inverse.

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