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The function f(x) is invertible, but the function g(x)=f(kx)  is not invertible. Find the sum of all possible values of k.

 Oct 6, 2018

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 #1
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\(\text{Assuming }f(x) \text{ is invertible }\forall x \in \mathbb{R} \\ \text{then }\forall k\neq 0, \exists y = k x \ni f^{-1}(y) = k x \\ \text{i.e. }f(kx) \text{ is invertible.}\\ \text{Thus it must be the only value of k that makes the original statement is valid is }\\ k=0 \\ \text{and the sum of this is just 0} \\ \text{It's true that }\exists f(0) \ni f^{-1}(f(0))=0 \text{ but }f(0\cdot x) \text{ is not invertible}\)

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 Oct 6, 2018
 #1
avatar+6248 
+1
Best Answer

\(\text{Assuming }f(x) \text{ is invertible }\forall x \in \mathbb{R} \\ \text{then }\forall k\neq 0, \exists y = k x \ni f^{-1}(y) = k x \\ \text{i.e. }f(kx) \text{ is invertible.}\\ \text{Thus it must be the only value of k that makes the original statement is valid is }\\ k=0 \\ \text{and the sum of this is just 0} \\ \text{It's true that }\exists f(0) \ni f^{-1}(f(0))=0 \text{ but }f(0\cdot x) \text{ is not invertible}\)

Rom Oct 6, 2018

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