Question

How do you find the inverse of `f(x) = ln(4 - 7x) + ln(-7 - 5x)`?

Answer 1

There is no inverse, because the quadratic formula yields two equations that returns two values for any given x and an inverse must not do this.

Explanation:

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

Substitute `f^-1(x)` everywhere there is an x:

`f(f^-1(x)) = ln(4 - 7f^-1(x)) + ln(-7 - 5f^-1(x))`

Use the property of inverses `f(f^-1(x)) = x`:

`x = ln(4 - 7f^-1(x)) + ln(-7 - 5f^-1(x))`

Use the identity `ln(a) + ln(b) = ln(ab)`

`x = ln((4 - 7f^-1(x)(-7 - 5f^-1(x)))`

Use the inverse of the natural logarithm on both sides:

`e^x = (4 - 7f^-1(x)(-7 - 5f^-1(x))`

Use the F.O.I.L. method to perform the multiplication:

`e^x = -28 - 20f^-1(x) + 49f^-1(x) + 35(f^-1(x))^2`

This is a quadratic.

`35(f^-1(x))^2 + 29f^-1(x) - 28 - e^x = 0`

At this point, we must declare that there is no inverse, because the quadratic formula yields two equations that returns two values for any given x and an inverse must not do this.

Let's proceed so that you can see the issue

Use the quadratic formula, `x = {-b +-sqrt(b^2 - 4(a)(c))}/{2a}`

`f^-1(x) = {-29 + sqrt(29^2 - 4(35)(-28-e^x))}/{2(35)}`

AND

`f^-1(x) = {-29 - sqrt(29^2 - 4(35)(-28-e^x))}/{2(35)}`

This is contrary to the purpose of an inverse function.

The purpose of an inverse function is, for any `y` returned from the `f(x)`, give you back a single value of `x` that created the `y`; this returns two values, therefore, no inverse.