Find inverse function of f(x) = x^2 (x in [1,4)]) please, help
y = x^2, solve for x then swap y and x
I need find : \[f:R\0\rightarrow R\\x\rightarrow \dfrac{1}{x^2}\] \[E=\{x \in R | 1\leq x\leq2\}\] find \(f^{-1}(E)\)
so that : let h(x) = 1/x--> h(E) =\(\{x\in R| 1\leq x\leq 2\}\) = [\(\dfrac{1}{2},1\)]
your R is suppose to be \( \sf \huge \mathbb{R} \)
and \(h^{-1}(x) = 1/x \) --> \(h^{-1}(E)=[1/2,1] \) also but \(f^{-1}(x) = (g(h(x))^{-1}= h^{-1}(g^{-1}(x) \)
Yes, that is Real
so that I need figure out what is \(g^{-1}(E)\)
first off, what is g^-(x)? is it \(g^{-1}(x) = \pm \sqrt{x}\)
On [1,4], it's just \(\sqrt{x}\). "Function" doesn't mean much when you deliberately write "\(\pm\)"
correct
it is important that you obtain the inverse function first, then test the conditions or range of values according to parameters
OK, thanks for the help. :)
Not quite. Rephrase. It is important that you obtain the inverse RELATION first, then test the conditions or range of values according to parameters SO THAT we have a function. There are two possible definitions for this f(x) on [1,4]. There is nothing sacred about choosing the positive one. Words mean things. Try to use them properly.
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