Monotonicity of tangent












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obviously for $a,b in ]frac{-pi}{2},frac{pi}{2}[$ the ordinary tangent map is strictly monotonic. Hence for $b geq a Rightarrow tan(b) geq tan(a)$. In the proof I try to understand it follows that $tan(]a,b[)=]tan(a),tan(b)[$. I need this since once proven I can conclude that tan maps open sets to open sets. It is clear that this statement is indeed true, but how to prove?



Thanks for your help!






real-analysis monotone-functions







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$endgroup$








  • 1




    $begingroup$
    Yes, for any continuous function $f:[a,b]rightarrow mathbb{R}$ you have $f([a,b]) = [min{f},max{f}]$ and in particular if $f$ is monotonic you know what the max and min are
    $endgroup$
    – user25959
    Dec 11 '18 at 16:20






  • 1




    $begingroup$
    It is a consequence of th intermediate value theorem.
    $endgroup$
    – Bernard
    Dec 11 '18 at 16:21










  • $begingroup$
    @user25959 Thanks, your first statement follows from the fact that any interval in $mathbb{R}$ is compact and since f is continous it follows that the image contains the minimum and maximum?
    $endgroup$
    – Michael Maier
    Dec 11 '18 at 16:24








  • 1




    $begingroup$
    The fact that the image is an interval follows from intermediate value theorem. That fact about maximum and minimum being attained is also true.
    $endgroup$
    – user25959
    Dec 11 '18 at 16:28
















0












$begingroup$


obviously for $a,b in ]frac{-pi}{2},frac{pi}{2}[$ the ordinary tangent map is strictly monotonic. Hence for $b geq a Rightarrow tan(b) geq tan(a)$. In the proof I try to understand it follows that $tan(]a,b[)=]tan(a),tan(b)[$. I need this since once proven I can conclude that tan maps open sets to open sets. It is clear that this statement is indeed true, but how to prove?



Thanks for your help!






real-analysis monotone-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, for any continuous function $f:[a,b]rightarrow mathbb{R}$ you have $f([a,b]) = [min{f},max{f}]$ and in particular if $f$ is monotonic you know what the max and min are
    $endgroup$
    – user25959
    Dec 11 '18 at 16:20






  • 1




    $begingroup$
    It is a consequence of th intermediate value theorem.
    $endgroup$
    – Bernard
    Dec 11 '18 at 16:21










  • $begingroup$
    @user25959 Thanks, your first statement follows from the fact that any interval in $mathbb{R}$ is compact and since f is continous it follows that the image contains the minimum and maximum?
    $endgroup$
    – Michael Maier
    Dec 11 '18 at 16:24








  • 1




    $begingroup$
    The fact that the image is an interval follows from intermediate value theorem. That fact about maximum and minimum being attained is also true.
    $endgroup$
    – user25959
    Dec 11 '18 at 16:28














0












0








0





$begingroup$


obviously for $a,b in ]frac{-pi}{2},frac{pi}{2}[$ the ordinary tangent map is strictly monotonic. Hence for $b geq a Rightarrow tan(b) geq tan(a)$. In the proof I try to understand it follows that $tan(]a,b[)=]tan(a),tan(b)[$. I need this since once proven I can conclude that tan maps open sets to open sets. It is clear that this statement is indeed true, but how to prove?



Thanks for your help!






real-analysis monotone-functions







share|cite|improve this question











$endgroup$




obviously for $a,b in ]frac{-pi}{2},frac{pi}{2}[$ the ordinary tangent map is strictly monotonic. Hence for $b geq a Rightarrow tan(b) geq tan(a)$. In the proof I try to understand it follows that $tan(]a,b[)=]tan(a),tan(b)[$. I need this since once proven I can conclude that tan maps open sets to open sets. It is clear that this statement is indeed true, but how to prove?



Thanks for your help!






real-analysis monotone-functions




real-analysis monotone-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 16:20









Bernard

120k740113




120k740113










asked Dec 11 '18 at 16:14









Michael MaierMichael Maier

859




859








  • 1




    $begingroup$
    Yes, for any continuous function $f:[a,b]rightarrow mathbb{R}$ you have $f([a,b]) = [min{f},max{f}]$ and in particular if $f$ is monotonic you know what the max and min are
    $endgroup$
    – user25959
    Dec 11 '18 at 16:20






  • 1




    $begingroup$
    It is a consequence of th intermediate value theorem.
    $endgroup$
    – Bernard
    Dec 11 '18 at 16:21










  • $begingroup$
    @user25959 Thanks, your first statement follows from the fact that any interval in $mathbb{R}$ is compact and since f is continous it follows that the image contains the minimum and maximum?
    $endgroup$
    – Michael Maier
    Dec 11 '18 at 16:24








  • 1




    $begingroup$
    The fact that the image is an interval follows from intermediate value theorem. That fact about maximum and minimum being attained is also true.
    $endgroup$
    – user25959
    Dec 11 '18 at 16:28














  • 1




    $begingroup$
    Yes, for any continuous function $f:[a,b]rightarrow mathbb{R}$ you have $f([a,b]) = [min{f},max{f}]$ and in particular if $f$ is monotonic you know what the max and min are
    $endgroup$
    – user25959
    Dec 11 '18 at 16:20






  • 1




    $begingroup$
    It is a consequence of th intermediate value theorem.
    $endgroup$
    – Bernard
    Dec 11 '18 at 16:21










  • $begingroup$
    @user25959 Thanks, your first statement follows from the fact that any interval in $mathbb{R}$ is compact and since f is continous it follows that the image contains the minimum and maximum?
    $endgroup$
    – Michael Maier
    Dec 11 '18 at 16:24








  • 1




    $begingroup$
    The fact that the image is an interval follows from intermediate value theorem. That fact about maximum and minimum being attained is also true.
    $endgroup$
    – user25959
    Dec 11 '18 at 16:28








1




1




$begingroup$
Yes, for any continuous function $f:[a,b]rightarrow mathbb{R}$ you have $f([a,b]) = [min{f},max{f}]$ and in particular if $f$ is monotonic you know what the max and min are
$endgroup$
– user25959
Dec 11 '18 at 16:20




$begingroup$
Yes, for any continuous function $f:[a,b]rightarrow mathbb{R}$ you have $f([a,b]) = [min{f},max{f}]$ and in particular if $f$ is monotonic you know what the max and min are
$endgroup$
– user25959
Dec 11 '18 at 16:20




1




1




$begingroup$
It is a consequence of th intermediate value theorem.
$endgroup$
– Bernard
Dec 11 '18 at 16:21




$begingroup$
It is a consequence of th intermediate value theorem.
$endgroup$
– Bernard
Dec 11 '18 at 16:21












$begingroup$
@user25959 Thanks, your first statement follows from the fact that any interval in $mathbb{R}$ is compact and since f is continous it follows that the image contains the minimum and maximum?
$endgroup$
– Michael Maier
Dec 11 '18 at 16:24






$begingroup$
@user25959 Thanks, your first statement follows from the fact that any interval in $mathbb{R}$ is compact and since f is continous it follows that the image contains the minimum and maximum?
$endgroup$
– Michael Maier
Dec 11 '18 at 16:24






1




1




$begingroup$
The fact that the image is an interval follows from intermediate value theorem. That fact about maximum and minimum being attained is also true.
$endgroup$
– user25959
Dec 11 '18 at 16:28




$begingroup$
The fact that the image is an interval follows from intermediate value theorem. That fact about maximum and minimum being attained is also true.
$endgroup$
– user25959
Dec 11 '18 at 16:28










1 Answer
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$begingroup$

$$xin (a,b)Longleftrightarrow a<x<b$$
$$Longleftrightarrow a<x and x<b$$
$$Longleftrightarrow tan(a)<tan(x) and tan(x)<tan(b)$$
which follows due to monotonicity of tangent function
$$Longleftrightarrow tan(x)in(tan(a),tan(b))$$



Hope it is helpful






share|cite|improve this answer









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    $begingroup$

    $$xin (a,b)Longleftrightarrow a<x<b$$
    $$Longleftrightarrow a<x and x<b$$
    $$Longleftrightarrow tan(a)<tan(x) and tan(x)<tan(b)$$
    which follows due to monotonicity of tangent function
    $$Longleftrightarrow tan(x)in(tan(a),tan(b))$$



    Hope it is helpful






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$xin (a,b)Longleftrightarrow a<x<b$$
      $$Longleftrightarrow a<x and x<b$$
      $$Longleftrightarrow tan(a)<tan(x) and tan(x)<tan(b)$$
      which follows due to monotonicity of tangent function
      $$Longleftrightarrow tan(x)in(tan(a),tan(b))$$



      Hope it is helpful






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$xin (a,b)Longleftrightarrow a<x<b$$
        $$Longleftrightarrow a<x and x<b$$
        $$Longleftrightarrow tan(a)<tan(x) and tan(x)<tan(b)$$
        which follows due to monotonicity of tangent function
        $$Longleftrightarrow tan(x)in(tan(a),tan(b))$$



        Hope it is helpful






        share|cite|improve this answer









        $endgroup$



        $$xin (a,b)Longleftrightarrow a<x<b$$
        $$Longleftrightarrow a<x and x<b$$
        $$Longleftrightarrow tan(a)<tan(x) and tan(x)<tan(b)$$
        which follows due to monotonicity of tangent function
        $$Longleftrightarrow tan(x)in(tan(a),tan(b))$$



        Hope it is helpful







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 16:22









        MartundMartund

        1,623213




        1,623213






























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