When is this function Differentiable?
$begingroup$
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!
calculus limits derivatives
edited Jan 8 at 19:49
Shubham Johri
4,759717
asked Jan 8 at 19:35
איתן לויאיתן לוי
284
$endgroup$
|
show 1 more comment
$begingroup$
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!
calculus limits derivatives
edited Jan 8 at 19:49
Shubham Johri
4,759717
asked Jan 8 at 19:35
איתן לויאיתן לוי
284
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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!
calculus limits derivatives
edited Jan 8 at 19:49
Shubham Johri
4,759717
asked Jan 8 at 19:35
איתן לויאיתן לוי
284
$endgroup$
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
calculus limits derivatives
edited Jan 8 at 19:49
Shubham Johri
4,759717
asked Jan 8 at 19:35
איתן לויאיתן לוי
284
edited Jan 8 at 19:49
Shubham Johri
4,759717
asked Jan 8 at 19:35
איתן לויאיתן לוי
284
edited Jan 8 at 19:49
Shubham Johri
4,759717
edited Jan 8 at 19:49
Shubham Johri
4,759717
edited Jan 8 at 19:49
Shubham Johri
4,759717
4,759717
asked Jan 8 at 19:35
איתן לויאיתן לוי
284
asked Jan 8 at 19:35
איתן לויאיתן לוי
284
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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:
answered Jan 8 at 19:59
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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:
answered Jan 8 at 19:59
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
add a comment |
$begingroup$
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:
answered Jan 8 at 19:59
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
add a comment |
$begingroup$
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:
answered Jan 8 at 19:59
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
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:
answered Jan 8 at 19:59
Mostafa AyazMostafa Ayaz
15.2k3939
answered Jan 8 at 19:59
Mostafa AyazMostafa Ayaz
15.2k3939
answered Jan 8 at 19:59
Mostafa AyazMostafa Ayaz
15.2k3939
answered Jan 8 at 19:59
Mostafa AyazMostafa Ayaz
15.2k3939
15.2k3939
add a comment |
add a comment |
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$begingroup$
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