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# HEEEELP

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If$$f(x)=\dfrac{x-3}{x-4}$$ , then for what value of $$x$$ is $$f^{-1}(x)$$ undefined?

Apr 19, 2022

#1
+9369
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We can actually find $$f^{-1}(x)$$ explicitly.

$$\begin{array}{rcl} f(x) &=& \dfrac{x - 3}{x - 4}\\ f\left(\color{red}f^{-1}(x)\right) &=& \dfrac{\color{red}f^{-1}(x)\color{black} - 3}{\color{red}f^{-1}(x)\color{black} - 4}\\ x &=& \dfrac{f^{-1}(x) - 3}{f^{-1}(x) - 4}\\ x(f^{-1}(x) - 4) &=& f^{-1}(x) - 3\\ xf^{-1}(x) - 4x &=& f^{-1}(x) - 3\\ (x - 1)f^{-1}(x) &=& 4x - 3\\ f^{-1}(x) &=& \dfrac{4x - 3}{x - 1} \end{array}$$

Therefore, $$f^{-1}(x)$$ is undefined when the denominator is 0, i.e., x - 1 = 0. Solving gives x = 1 as the answer.

Apr 19, 2022
#2
+122
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That is correct i believe

Kakashi  Apr 19, 2022
#3
+9369
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Yes, that's true. I checked with Wolfram Alpha.

MaxWong  Apr 19, 2022