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4.Given the function
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\[f:x \rightarrow \frac{ 2x }{ x-1 },x1\]
Doesn't really make sense.
yeagh
b) state the value of x such that f^-1 does not exist.
so you cannot divide by 0 right
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so we need to make denominator 0
x-1=0 x?
x=1
so it will be 2? @AlexandervonHumboldt2
yes so the function doesnt exist where x=1
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Thnx @AlexandervonHumboldt2
Yeah, that seems correct.
txh you too
You will find a vertical asymptote at x=1. That is where the function dne.
Thnx @Jhannybean
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Np
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