When is this function Differentiable?












1












$begingroup$


I was given this function:



$$f(x)=begin{cases}displaystyle|x|^pcosBig(fracpi{|x|^q}Big),&xne0\0,&x=0end{cases}$$



And was asked to find for what $p, q>0$ it is differentiable at $x=0$.



First I saw it is continuous when $p>0, q>0$.



Now, I tried to see if the limit for $f'(x)$ exists at $x=0$. This function is even so I looked at the right side only.



$$lim_{xto0^+}frac{f(x)-f(0)}{x-0}=lim_{xto0^+}frac{x^pcosBig(displaystylefracpi{x^q}Big)}x=lim_{xto0^+}x^{p-1}cosBig(fracpi{x^q}Big)$$



I get that this limit exists when $p>1$, and for all $q>0$, but looking at the graph online it doesn't seem to be right. What am I doing wrong here (if anything)?



Thanks a lot!










share|cite|improve this question











$endgroup$












  • $begingroup$
    Why did you remove the limit?
    $endgroup$
    – EuxhenH
    Jan 8 at 19:38










  • $begingroup$
    Just because it takes like an hour to write the limit each time in overleaf. Assume it's still there.
    $endgroup$
    – איתן לוי
    Jan 8 at 19:40










  • $begingroup$
    lol an hour????
    $endgroup$
    – Randall
    Jan 8 at 19:45










  • $begingroup$
    When you are a complete newbie it takes a few minutes. Is this such a problem?
    $endgroup$
    – איתן לוי
    Jan 8 at 19:49










  • $begingroup$
    You have solved it correctly. Graphical aids are often inaccurate for plots like these. How were you able to draw any conclusion for $f$ near $0$ using its graph?
    $endgroup$
    – Shubham Johri
    Jan 8 at 19:53


















1












$begingroup$


I was given this function:



$$f(x)=begin{cases}displaystyle|x|^pcosBig(fracpi{|x|^q}Big),&xne0\0,&x=0end{cases}$$



And was asked to find for what $p, q>0$ it is differentiable at $x=0$.



First I saw it is continuous when $p>0, q>0$.



Now, I tried to see if the limit for $f'(x)$ exists at $x=0$. This function is even so I looked at the right side only.



$$lim_{xto0^+}frac{f(x)-f(0)}{x-0}=lim_{xto0^+}frac{x^pcosBig(displaystylefracpi{x^q}Big)}x=lim_{xto0^+}x^{p-1}cosBig(fracpi{x^q}Big)$$



I get that this limit exists when $p>1$, and for all $q>0$, but looking at the graph online it doesn't seem to be right. What am I doing wrong here (if anything)?



Thanks a lot!










share|cite|improve this question











$endgroup$












  • $begingroup$
    Why did you remove the limit?
    $endgroup$
    – EuxhenH
    Jan 8 at 19:38










  • $begingroup$
    Just because it takes like an hour to write the limit each time in overleaf. Assume it's still there.
    $endgroup$
    – איתן לוי
    Jan 8 at 19:40










  • $begingroup$
    lol an hour????
    $endgroup$
    – Randall
    Jan 8 at 19:45










  • $begingroup$
    When you are a complete newbie it takes a few minutes. Is this such a problem?
    $endgroup$
    – איתן לוי
    Jan 8 at 19:49










  • $begingroup$
    You have solved it correctly. Graphical aids are often inaccurate for plots like these. How were you able to draw any conclusion for $f$ near $0$ using its graph?
    $endgroup$
    – Shubham Johri
    Jan 8 at 19:53
















1












1








1





$begingroup$


I was given this function:



$$f(x)=begin{cases}displaystyle|x|^pcosBig(fracpi{|x|^q}Big),&xne0\0,&x=0end{cases}$$



And was asked to find for what $p, q>0$ it is differentiable at $x=0$.



First I saw it is continuous when $p>0, q>0$.



Now, I tried to see if the limit for $f'(x)$ exists at $x=0$. This function is even so I looked at the right side only.



$$lim_{xto0^+}frac{f(x)-f(0)}{x-0}=lim_{xto0^+}frac{x^pcosBig(displaystylefracpi{x^q}Big)}x=lim_{xto0^+}x^{p-1}cosBig(fracpi{x^q}Big)$$



I get that this limit exists when $p>1$, and for all $q>0$, but looking at the graph online it doesn't seem to be right. What am I doing wrong here (if anything)?



Thanks a lot!










share|cite|improve this question











$endgroup$




I was given this function:



$$f(x)=begin{cases}displaystyle|x|^pcosBig(fracpi{|x|^q}Big),&xne0\0,&x=0end{cases}$$



And was asked to find for what $p, q>0$ it is differentiable at $x=0$.



First I saw it is continuous when $p>0, q>0$.



Now, I tried to see if the limit for $f'(x)$ exists at $x=0$. This function is even so I looked at the right side only.



$$lim_{xto0^+}frac{f(x)-f(0)}{x-0}=lim_{xto0^+}frac{x^pcosBig(displaystylefracpi{x^q}Big)}x=lim_{xto0^+}x^{p-1}cosBig(fracpi{x^q}Big)$$



I get that this limit exists when $p>1$, and for all $q>0$, but looking at the graph online it doesn't seem to be right. What am I doing wrong here (if anything)?



Thanks a lot!







calculus limits derivatives






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 19:49









Shubham Johri

5,187717




5,187717










asked Jan 8 at 19:35









איתן לויאיתן לוי

284




284












  • $begingroup$
    Why did you remove the limit?
    $endgroup$
    – EuxhenH
    Jan 8 at 19:38










  • $begingroup$
    Just because it takes like an hour to write the limit each time in overleaf. Assume it's still there.
    $endgroup$
    – איתן לוי
    Jan 8 at 19:40










  • $begingroup$
    lol an hour????
    $endgroup$
    – Randall
    Jan 8 at 19:45










  • $begingroup$
    When you are a complete newbie it takes a few minutes. Is this such a problem?
    $endgroup$
    – איתן לוי
    Jan 8 at 19:49










  • $begingroup$
    You have solved it correctly. Graphical aids are often inaccurate for plots like these. How were you able to draw any conclusion for $f$ near $0$ using its graph?
    $endgroup$
    – Shubham Johri
    Jan 8 at 19:53




















  • $begingroup$
    Why did you remove the limit?
    $endgroup$
    – EuxhenH
    Jan 8 at 19:38










  • $begingroup$
    Just because it takes like an hour to write the limit each time in overleaf. Assume it's still there.
    $endgroup$
    – איתן לוי
    Jan 8 at 19:40










  • $begingroup$
    lol an hour????
    $endgroup$
    – Randall
    Jan 8 at 19:45










  • $begingroup$
    When you are a complete newbie it takes a few minutes. Is this such a problem?
    $endgroup$
    – איתן לוי
    Jan 8 at 19:49










  • $begingroup$
    You have solved it correctly. Graphical aids are often inaccurate for plots like these. How were you able to draw any conclusion for $f$ near $0$ using its graph?
    $endgroup$
    – Shubham Johri
    Jan 8 at 19:53


















$begingroup$
Why did you remove the limit?
$endgroup$
– EuxhenH
Jan 8 at 19:38




$begingroup$
Why did you remove the limit?
$endgroup$
– EuxhenH
Jan 8 at 19:38












$begingroup$
Just because it takes like an hour to write the limit each time in overleaf. Assume it's still there.
$endgroup$
– איתן לוי
Jan 8 at 19:40




$begingroup$
Just because it takes like an hour to write the limit each time in overleaf. Assume it's still there.
$endgroup$
– איתן לוי
Jan 8 at 19:40












$begingroup$
lol an hour????
$endgroup$
– Randall
Jan 8 at 19:45




$begingroup$
lol an hour????
$endgroup$
– Randall
Jan 8 at 19:45












$begingroup$
When you are a complete newbie it takes a few minutes. Is this such a problem?
$endgroup$
– איתן לוי
Jan 8 at 19:49




$begingroup$
When you are a complete newbie it takes a few minutes. Is this such a problem?
$endgroup$
– איתן לוי
Jan 8 at 19:49












$begingroup$
You have solved it correctly. Graphical aids are often inaccurate for plots like these. How were you able to draw any conclusion for $f$ near $0$ using its graph?
$endgroup$
– Shubham Johri
Jan 8 at 19:53






$begingroup$
You have solved it correctly. Graphical aids are often inaccurate for plots like these. How were you able to draw any conclusion for $f$ near $0$ using its graph?
$endgroup$
– Shubham Johri
Jan 8 at 19:53












1 Answer
1






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oldest

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0












$begingroup$

Your conclusion is right for $p>1$ and all values of $q>0$. The reason why you don't observe so on the graph is the the oscillation of the function increases around $x=0$ so it's indistinguishable to see whether the function is differentiable in $x=0$ or not. Also the function has no continuous derivative in $x=0$ for $0<ple 1$. The figure below shows why:



enter image description here






share|cite|improve this answer









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    1 Answer
    1






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    active

    oldest

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    0












    $begingroup$

    Your conclusion is right for $p>1$ and all values of $q>0$. The reason why you don't observe so on the graph is the the oscillation of the function increases around $x=0$ so it's indistinguishable to see whether the function is differentiable in $x=0$ or not. Also the function has no continuous derivative in $x=0$ for $0<ple 1$. The figure below shows why:



    enter image description here






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your conclusion is right for $p>1$ and all values of $q>0$. The reason why you don't observe so on the graph is the the oscillation of the function increases around $x=0$ so it's indistinguishable to see whether the function is differentiable in $x=0$ or not. Also the function has no continuous derivative in $x=0$ for $0<ple 1$. The figure below shows why:



      enter image description here






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your conclusion is right for $p>1$ and all values of $q>0$. The reason why you don't observe so on the graph is the the oscillation of the function increases around $x=0$ so it's indistinguishable to see whether the function is differentiable in $x=0$ or not. Also the function has no continuous derivative in $x=0$ for $0<ple 1$. The figure below shows why:



        enter image description here






        share|cite|improve this answer









        $endgroup$



        Your conclusion is right for $p>1$ and all values of $q>0$. The reason why you don't observe so on the graph is the the oscillation of the function increases around $x=0$ so it's indistinguishable to see whether the function is differentiable in $x=0$ or not. Also the function has no continuous derivative in $x=0$ for $0<ple 1$. The figure below shows why:



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 19:59









        Mostafa AyazMostafa Ayaz

        15.6k3939




        15.6k3939






























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