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I've stumbled across this seemingly easy problem. Prove that for every continuous bijective rising function, this is true: $$(c\cdot f)^{-1}(x)=f^{-1}(\frac{x}{c})$$ where c $\in$ (0,$\infty)$.

My question is this - I know that this is true: $$(c\cdot f)(x)=c\cdot f(x)$$ But I have no idea how this equation behaves when dealing with inverse functions. Suppose I have a rising invertible function like this: $$f(x)=x$$ Let c=2. What does this -> $(c\cdot f)^{-1}(x)$ <- return? Or rather, how do I read the notation? Do I first invert the function? (and get the same function?) and then multiply by two (getting y=2x) or do I multiply it by two (and make it 2x) and then invert (getting y=0.5x)?

Edit: Another question of mine is: What does multiplying a function by a constant even mean? If I had a function f(x)=x+1, then (2*f)(x) is 2x+1 or 2x+2?

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    $\begingroup$ "What does multiplying a function by a constant even mean?" You just gave the definition in "I know that this is true..." $\endgroup$ – user9464 Feb 13 at 23:44
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Hint: try to prove this: if $f : \Bbb{R} \to \Bbb{R}$ is monotone increasing and continuous, then for any $c$, $f^{-1}(c)$ is well-defined and $f^{-1}(c) = \inf\{x \mid f(x) \ge c\}$.

As for the other question you added: the result of multiplying a function $f$ by a constant $c$ is the function $x \mapsto c\cdot f(x)$. It's the composition of $f$ with multiplication by $c$ - as in the correct statement in the second displayed equation in your question.

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