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