# Solving inverse function equations

A function f is defined by $$f(x) = \displaystyle\frac{2x-3}{x-1}, x≠1$$. Solve the equation |$$f^{-1}(x)$$| = 1 + $$f$$-1$$(x)$$.

I first found out the inverse and equated for the left hand side the negative of the inverse and then solved. However, I got the wrong answer and was unsure why.

• " However, I got the wrong answer and was unsure why." And how would we know why you got it wrong? We don't know what you did. Commented Dec 8, 2019 at 6:42
• I got x as -8 and 2.
– V11
Commented Dec 8, 2019 at 6:44
• HOW did you get a $-8$ and a $2$? We can't help you if we don't know what you did. Commented Dec 8, 2019 at 6:48
• (3-x)(x-2) = (-x+2)(2x-5) and I simplified this further to get x^2 - 2x + 8x - 16 = 0 and got -8 and 2 as the answer.
– V11
Commented Dec 8, 2019 at 6:53
• "I first found out the inverse" And what was it? "and equated for the left hand side the negative of the inverse" why? $|A| \ne -A$ unless $A \le 0$. IS $f(x) \le 0$. "and then solved" solved what equation? Commented Dec 8, 2019 at 16:49

Finding the inverse is actually unnecessary. The only solution to the equation to

$$|z| = 1 + z$$

is $$z = -\frac{1}{2}$$ (this is because the only way absolute value gives you a different number from the input is when the input is negative).

In other words we have that

$$f^{-1}(x) = -\frac{1}{2} \implies x = f\left(-\frac{1}{2}\right) = \frac{8}{3}$$

• Oh right, I get it now. Thank you!
– V11
Commented Dec 8, 2019 at 6:41