# Function inverse notation

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?

• "What does multiplying a function by a constant even mean?" You just gave the definition in "I know that this is true..." – user9464 Feb 13 at 23:44

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.