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Question 1 : What's the difference between $(f^{-1}of)(x)$ and$(fof^{-1})(y)$ ? Also can we say $(f^{-1}of)(x) = (fof^{-1})(x) = x$ ?

Question 2 : Why these notations are different ? $\arctan(x)$ , $arc(\frac{sin(x)}{cos(x)})$ and $\frac{\arcsin(x)}{\arccos(x)}$

Question 3 : How we can find intersection points of $f(x)$ and $f^{-1}(x)$ or how we can solve $f(x) = f^{-1}(x)$ ?

My try : I really don't know any trustable source for finding the answers.

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    Try putting these as three separate questions. That way you get three answers dedicated to the three questions. – 2017-02-03
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    For Question 2: It is not true that $\arctan(x) = \arcsin(x) / \arccos(x)$. A possible source of confusion: the notation $\sin^{-1} (x)$ means $\arcsin(x)$, with the -1 denoting a functional inverse, not a reciprocal: $\sin^{-1}(x) \neq 1/\sin(x) = (\sin(x))^{-1}$. On the other hand, $\sin^2(x) = (\sin(x))^2 $, $\sin^3(x) = (\sin(x))^3 $, _etc._. This is an unfortunate notational convention. Also, we don't write $ \mathrm{arc}(\sin(x) / \cos(x)) $, because $\mathrm{arc}$ is part of the names of the functions $\arcsin$, _etc._, not a function in its own right. – 2017-08-17

2 Answers 2

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Case $1$: For functions $f:X → Y$ and $f^{−1}:Y → X$, $${\displaystyle f^{-1}\circ f=\mathrm {id} _{X}} \text { and } {\displaystyle f\circ f^{-1}=\mathrm {id} _{Y}.}$$ where $\mathrm {id}_x $ is the identity function on the set $X$. So there is no difference.

Case $2$: It's true that $\frac {\sin x}{\cos x}=\tan x $, but the functions $\arcsin ⁡x, \arccos⁡ x$ and $\arctan⁡ x$ are the inverse functions of the former. If you think about it, it actually makes more sense that if some relation is true for a set of functions, it will not be true for their inverse functions.

Case $3$: Has been discussed many times on the site, for example see here.

Hope it helps.

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    But I think $id_x$ and $id_y$ has different – 2017-02-03
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In general we do not have $(f^{-1}of)(x) = (fof^{-1})(x) = x$

Example: Let $A=\{1,2\}$, $B=\{3,4\}$ and let $f :A \to B$ be defined by

$f(1)=3$, $f(2)=4$.

Then $f$ is bijective and $f^{-1}:B \to A$ is given by

$f^{-1}(3)=1$ and $f^{-1}(4)=2$.

We have $(f^{-1}of)(a) = a$ for all $a \in A$ and $(fof^{-1})(b) = b$ for all $b \in B$ .

Observe that $A \cap B= \emptyset$.

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    In my book was said that "$fof^{-1} = f^{-1}of$" . Is it true ? – 2017-02-03
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    No. My example shows that this is not true ! In my example $fof^{-1} $ is defined on $B$, but $f ^{-1}of$ is defined on $A$. Since $A \ne B$, $fof^{-1}$ and $f^{-1}of$ are different. – 2017-02-03
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    @S.H.W: Did your book specify that $f:X \to X$ (i.e., the domain and target/codomain are the same set)? If so, that's an important detail in your question, and your book is correct. – 2017-02-03
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    @AndrewD.Hwang In my book was said "$x \in D_{f^{-1}} : fof^{-1}(x) = x$ and $x \in D_{f} : f^{-1}of(x) = x $" – 2017-02-03
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    Yes, that's correct. But in general $D_f \ne D_{f^{-1}}$ (see my example). Hence $fof^{-1} \ne f^{-1}of$ – 2017-02-03
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    So , this phrase in the book is wrong – 2017-02-03
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    Your book says: for $x \in D_{f^{-1}} : fof^{-1}(x) = x$ and for $x \in D_{f} : f^{-1}of(x) = x$ . That's correct. It would be better to say: for $b \in D_{f^{-1}} : fof^{-1}(b) = b$ and for $a \in D_{f} : f^{-1}of(a) = a$ – 2017-02-03
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    Okay , thank you a lot for your help . Can you provide a method for solving $f(x) = f^{-1}(x)$ ? I read many answers in this site but I can't understand it completely . – 2017-02-03