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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.

Zera
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V11
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  • " 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. – fleablood Dec 08 '19 at 06:42
  • I got x as -8 and 2. – V11 Dec 08 '19 at 06:44
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    HOW did you get a $-8$ and a $2$? We can't help you if we don't know what you did. – fleablood Dec 08 '19 at 06:48
  • (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 Dec 08 '19 at 06:53
  • "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? – fleablood Dec 08 '19 at 16:49
  • Okay..... if $f^{-1}(x) = \frac {x-3}{x-2}$ and if you figured for $|f^{-1}(x)| = 1 +f^{-1}(x)$ we can't have $|f^{-1}(x)|= f^{-1}(x)|$ so $f^{-1}(x) < 0$.... then your right hand side ought to have been $|f^{-1}(x)|=\frac {3-x}{x-2}$ and $1+f^{-1}(x)=1+\frac{x-3}{x-2}=\frac{(x-3)+(x-2)}{x-2}= \frac {2x-5}{x-2}$ so your equation should have been $\frac {3-x}{x-2}=\frac {2x-5}{x-2}$ and so $3-x = 2x-5$. – fleablood Dec 08 '19 at 17:13

1 Answers1

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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

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

Ninad Munshi
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