How to show that the function $f(x) = sin x + tan x$ on $(-pi/2, pi/2)$ has an inverse












1












$begingroup$


Given that $f(x)=sin x+tan x$ on $(-pi/2,pi/2)$. Show that it has an inverse function. Find the derivative of the inverse of $f(x)$ at $x=0$.



What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?






calculus algebra-precalculus







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$endgroup$








  • 1




    $begingroup$
    Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
    $endgroup$
    – Deepak
    Dec 8 '18 at 9:44






  • 2




    $begingroup$
    In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
    $endgroup$
    – achille hui
    Dec 8 '18 at 9:49










  • $begingroup$
    @achille hui I think that we need not only injection but bijection to have an inverse for a function.
    $endgroup$
    – hamza boulahia
    Dec 8 '18 at 10:01










  • $begingroup$
    @hamzaboulahia Injection is sufficient.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:02










  • $begingroup$
    @hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
    $endgroup$
    – achille hui
    Dec 8 '18 at 10:05
















1












$begingroup$


Given that $f(x)=sin x+tan x$ on $(-pi/2,pi/2)$. Show that it has an inverse function. Find the derivative of the inverse of $f(x)$ at $x=0$.



What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?






calculus algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
    $endgroup$
    – Deepak
    Dec 8 '18 at 9:44






  • 2




    $begingroup$
    In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
    $endgroup$
    – achille hui
    Dec 8 '18 at 9:49










  • $begingroup$
    @achille hui I think that we need not only injection but bijection to have an inverse for a function.
    $endgroup$
    – hamza boulahia
    Dec 8 '18 at 10:01










  • $begingroup$
    @hamzaboulahia Injection is sufficient.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:02










  • $begingroup$
    @hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
    $endgroup$
    – achille hui
    Dec 8 '18 at 10:05














1












1








1





$begingroup$


Given that $f(x)=sin x+tan x$ on $(-pi/2,pi/2)$. Show that it has an inverse function. Find the derivative of the inverse of $f(x)$ at $x=0$.



What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?






calculus algebra-precalculus







share|cite|improve this question











$endgroup$




Given that $f(x)=sin x+tan x$ on $(-pi/2,pi/2)$. Show that it has an inverse function. Find the derivative of the inverse of $f(x)$ at $x=0$.



What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?






calculus algebra-precalculus




calculus algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 11:03









N. F. Taussig

44.1k93356




44.1k93356










asked Dec 8 '18 at 9:40









HouseBTHouseBT

154




154








  • 1




    $begingroup$
    Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
    $endgroup$
    – Deepak
    Dec 8 '18 at 9:44






  • 2




    $begingroup$
    In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
    $endgroup$
    – achille hui
    Dec 8 '18 at 9:49










  • $begingroup$
    @achille hui I think that we need not only injection but bijection to have an inverse for a function.
    $endgroup$
    – hamza boulahia
    Dec 8 '18 at 10:01










  • $begingroup$
    @hamzaboulahia Injection is sufficient.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:02










  • $begingroup$
    @hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
    $endgroup$
    – achille hui
    Dec 8 '18 at 10:05














  • 1




    $begingroup$
    Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
    $endgroup$
    – Deepak
    Dec 8 '18 at 9:44






  • 2




    $begingroup$
    In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
    $endgroup$
    – achille hui
    Dec 8 '18 at 9:49










  • $begingroup$
    @achille hui I think that we need not only injection but bijection to have an inverse for a function.
    $endgroup$
    – hamza boulahia
    Dec 8 '18 at 10:01










  • $begingroup$
    @hamzaboulahia Injection is sufficient.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:02










  • $begingroup$
    @hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
    $endgroup$
    – achille hui
    Dec 8 '18 at 10:05








1




1




$begingroup$
Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
$endgroup$
– Deepak
Dec 8 '18 at 9:44




$begingroup$
Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
$endgroup$
– Deepak
Dec 8 '18 at 9:44




2




2




$begingroup$
In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
$endgroup$
– achille hui
Dec 8 '18 at 9:49




$begingroup$
In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
$endgroup$
– achille hui
Dec 8 '18 at 9:49












$begingroup$
@achille hui I think that we need not only injection but bijection to have an inverse for a function.
$endgroup$
– hamza boulahia
Dec 8 '18 at 10:01




$begingroup$
@achille hui I think that we need not only injection but bijection to have an inverse for a function.
$endgroup$
– hamza boulahia
Dec 8 '18 at 10:01












$begingroup$
@hamzaboulahia Injection is sufficient.
$endgroup$
– Rebellos
Dec 8 '18 at 10:02




$begingroup$
@hamzaboulahia Injection is sufficient.
$endgroup$
– Rebellos
Dec 8 '18 at 10:02












$begingroup$
@hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
$endgroup$
– achille hui
Dec 8 '18 at 10:05




$begingroup$
@hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
$endgroup$
– achille hui
Dec 8 '18 at 10:05










2 Answers
2






active

oldest

votes


















1












$begingroup$

In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



But, consider the following standard Theorem proved in pre-calculus courses :




Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
$$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
Proof :



Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
By definition, it is :
$$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




Can you now use that theorem to find the derivative of the inverse at $x=0$ ?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:15












  • $begingroup$
    Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:22










  • $begingroup$
    I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:28












  • $begingroup$
    What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:30










  • $begingroup$
    Okay now i understand my mistake and the right path to follow thank you for your time ^^
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:37





















1












$begingroup$

$star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



$star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



    But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



    Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



    But, consider the following standard Theorem proved in pre-calculus courses :




    Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
    $$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
    Proof :



    Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
    By definition, it is :
    $$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




    Can you now use that theorem to find the derivative of the inverse at $x=0$ ?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:15












    • $begingroup$
      Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:22










    • $begingroup$
      I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:28












    • $begingroup$
      What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:30










    • $begingroup$
      Okay now i understand my mistake and the right path to follow thank you for your time ^^
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:37


















    1












    $begingroup$

    In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



    But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



    Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



    But, consider the following standard Theorem proved in pre-calculus courses :




    Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
    $$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
    Proof :



    Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
    By definition, it is :
    $$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




    Can you now use that theorem to find the derivative of the inverse at $x=0$ ?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:15












    • $begingroup$
      Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:22










    • $begingroup$
      I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:28












    • $begingroup$
      What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:30










    • $begingroup$
      Okay now i understand my mistake and the right path to follow thank you for your time ^^
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:37
















    1












    1








    1





    $begingroup$

    In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



    But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



    Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



    But, consider the following standard Theorem proved in pre-calculus courses :




    Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
    $$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
    Proof :



    Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
    By definition, it is :
    $$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




    Can you now use that theorem to find the derivative of the inverse at $x=0$ ?






    share|cite|improve this answer











    $endgroup$



    In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



    But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



    Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



    But, consider the following standard Theorem proved in pre-calculus courses :




    Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
    $$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
    Proof :



    Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
    By definition, it is :
    $$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




    Can you now use that theorem to find the derivative of the inverse at $x=0$ ?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 8 '18 at 10:08

























    answered Dec 8 '18 at 10:00









    RebellosRebellos

    14.6k31247




    14.6k31247












    • $begingroup$
      Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:15












    • $begingroup$
      Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:22










    • $begingroup$
      I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:28












    • $begingroup$
      What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:30










    • $begingroup$
      Okay now i understand my mistake and the right path to follow thank you for your time ^^
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:37




















    • $begingroup$
      Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:15












    • $begingroup$
      Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:22










    • $begingroup$
      I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:28












    • $begingroup$
      What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:30










    • $begingroup$
      Okay now i understand my mistake and the right path to follow thank you for your time ^^
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:37


















    $begingroup$
    Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:15






    $begingroup$
    Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:15














    $begingroup$
    Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:22




    $begingroup$
    Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:22












    $begingroup$
    I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:28






    $begingroup$
    I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:28














    $begingroup$
    What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:30




    $begingroup$
    What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:30












    $begingroup$
    Okay now i understand my mistake and the right path to follow thank you for your time ^^
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:37






    $begingroup$
    Okay now i understand my mistake and the right path to follow thank you for your time ^^
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:37













    1












    $begingroup$

    $star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



    $star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
    But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



      $star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
      But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



        $star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
        But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$






        share|cite|improve this answer









        $endgroup$



        $star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



        $star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
        But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 9:59









        hamza boulahiahamza boulahia

        989419




        989419






























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