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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
avatar+9459 
+2

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
avatar+115 
-4

That is correct i believe 

Kakashi  Apr 19, 2022
 #3
avatar+9459 
+2

Yes, that's true. I checked with Wolfram Alpha.

MaxWong  Apr 19, 2022

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