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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.

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  • $\begingroup$ " 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. $\endgroup$
    – fleablood
    Commented Dec 8, 2019 at 6:42
  • $\begingroup$ I got x as -8 and 2. $\endgroup$
    – V11
    Commented Dec 8, 2019 at 6:44
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    $\begingroup$ HOW did you get a $-8$ and a $2$? We can't help you if we don't know what you did. $\endgroup$
    – fleablood
    Commented Dec 8, 2019 at 6:48
  • $\begingroup$ (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. $\endgroup$
    – V11
    Commented Dec 8, 2019 at 6:53
  • $\begingroup$ "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? $\endgroup$
    – fleablood
    Commented Dec 8, 2019 at 16:49

1 Answer 1

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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

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

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

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