How to check the differentiability of $f(x) = x|x|$ at $x_0 = 0$?
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How do I check the differentiability of a $f(x) = x|x|$ at $x_0=0$?
I used the definition of a derivative and came up with
begin{align}
f^prime &= lim_{xto0}dfrac{f(x)-f(x)}{x-0}\
&= lim_{xto0}dfrac{x|x|-0|0|}{x-0}\
&= lim_{xto0}dfrac{x|x|}{x}\
&= lim_{xto0}|x|
end{align}
I know $f(x)=|x|$ is not differentiable at $0$. However, would this be? If one plugs in $0$ to $|x|$, one would get $0$, which is finite. But, if one takes the the limit from $-infty$ and $+infty$, the limits wouldn't exist.
calculus derivatives
asked Nov 17 at 21:24
kaisa
717
add a comment |
up vote
0
down vote
favorite
How do I check the differentiability of a $f(x) = x|x|$ at $x_0=0$?
I used the definition of a derivative and came up with
begin{align}
f^prime &= lim_{xto0}dfrac{f(x)-f(x)}{x-0}\
&= lim_{xto0}dfrac{x|x|-0|0|}{x-0}\
&= lim_{xto0}dfrac{x|x|}{x}\
&= lim_{xto0}|x|
end{align}
I know $f(x)=|x|$ is not differentiable at $0$. However, would this be? If one plugs in $0$ to $|x|$, one would get $0$, which is finite. But, if one takes the the limit from $-infty$ and $+infty$, the limits wouldn't exist.
calculus derivatives
asked Nov 17 at 21:24
kaisa
717
1
Why would you take the limit to $;pminfty;$ at all? The limit is when $;xto0;$ ...and that's all!
– DonAntonio
Nov 17 at 21:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I check the differentiability of a $f(x) = x|x|$ at $x_0=0$?
I used the definition of a derivative and came up with
begin{align}
f^prime &= lim_{xto0}dfrac{f(x)-f(x)}{x-0}\
&= lim_{xto0}dfrac{x|x|-0|0|}{x-0}\
&= lim_{xto0}dfrac{x|x|}{x}\
&= lim_{xto0}|x|
end{align}
I know $f(x)=|x|$ is not differentiable at $0$. However, would this be? If one plugs in $0$ to $|x|$, one would get $0$, which is finite. But, if one takes the the limit from $-infty$ and $+infty$, the limits wouldn't exist.
calculus derivatives
asked Nov 17 at 21:24
kaisa
717
How do I check the differentiability of a $f(x) = x|x|$ at $x_0=0$?
I used the definition of a derivative and came up with
begin{align}
f^prime &= lim_{xto0}dfrac{f(x)-f(x)}{x-0}\
&= lim_{xto0}dfrac{x|x|-0|0|}{x-0}\
&= lim_{xto0}dfrac{x|x|}{x}\
&= lim_{xto0}|x|
end{align}
I know $f(x)=|x|$ is not differentiable at $0$. However, would this be? If one plugs in $0$ to $|x|$, one would get $0$, which is finite. But, if one takes the the limit from $-infty$ and $+infty$, the limits wouldn't exist.
calculus derivatives
calculus derivatives
asked Nov 17 at 21:24
kaisa
717
asked Nov 17 at 21:24
kaisa
717
asked Nov 17 at 21:24
kaisa
717
asked Nov 17 at 21:24
kaisa
717
asked Nov 17 at 21:24
kaisa
717
717
1
Why would you take the limit to $;pminfty;$ at all? The limit is when $;xto0;$ ...and that's all!
– DonAntonio
Nov 17 at 21:28
add a comment |
1
Why would you take the limit to $;pminfty;$ at all? The limit is when $;xto0;$ ...and that's all!
– DonAntonio
Nov 17 at 21:28
1
1
Why would you take the limit to $;pminfty;$ at all? The limit is when $;xto0;$ ...and that's all!
– DonAntonio
Nov 17 at 21:28
Why would you take the limit to $;pminfty;$ at all? The limit is when $;xto0;$ ...and that's all!
– DonAntonio
Nov 17 at 21:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Why are you mentioning $pminfty$? What you did is correct. Since $lim_{xto0}lvert xrvert=0$, $f'(0)=0$.
answered Nov 17 at 21:26
José Carlos Santos
143k20112208
Would $$lim_{xto0^+} |x|=lim_{xto0^-} |x|?$$ That is why I mentioned coming from either infinity to 0.
– kaisa
Nov 17 at 21:29
Both of those limits are equal to $0$. In other words, $lim_{xto0}lvert xrvert=0$.
– José Carlos Santos
Nov 17 at 21:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Why are you mentioning $pminfty$? What you did is correct. Since $lim_{xto0}lvert xrvert=0$, $f'(0)=0$.
answered Nov 17 at 21:26
José Carlos Santos
143k20112208
Would $$lim_{xto0^+} |x|=lim_{xto0^-} |x|?$$ That is why I mentioned coming from either infinity to 0.
– kaisa
Nov 17 at 21:29
Both of those limits are equal to $0$. In other words, $lim_{xto0}lvert xrvert=0$.
– José Carlos Santos
Nov 17 at 21:32
add a comment |
up vote
2
down vote
accepted
Why are you mentioning $pminfty$? What you did is correct. Since $lim_{xto0}lvert xrvert=0$, $f'(0)=0$.
answered Nov 17 at 21:26
José Carlos Santos
143k20112208
Would $$lim_{xto0^+} |x|=lim_{xto0^-} |x|?$$ That is why I mentioned coming from either infinity to 0.
– kaisa
Nov 17 at 21:29
Both of those limits are equal to $0$. In other words, $lim_{xto0}lvert xrvert=0$.
– José Carlos Santos
Nov 17 at 21:32
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Why are you mentioning $pminfty$? What you did is correct. Since $lim_{xto0}lvert xrvert=0$, $f'(0)=0$.
answered Nov 17 at 21:26
José Carlos Santos
143k20112208
Why are you mentioning $pminfty$? What you did is correct. Since $lim_{xto0}lvert xrvert=0$, $f'(0)=0$.
answered Nov 17 at 21:26
José Carlos Santos
143k20112208
answered Nov 17 at 21:26
José Carlos Santos
143k20112208
answered Nov 17 at 21:26
José Carlos Santos
143k20112208
answered Nov 17 at 21:26
José Carlos Santos
143k20112208
143k20112208
Would $$lim_{xto0^+} |x|=lim_{xto0^-} |x|?$$ That is why I mentioned coming from either infinity to 0.
– kaisa
Nov 17 at 21:29
Both of those limits are equal to $0$. In other words, $lim_{xto0}lvert xrvert=0$.
– José Carlos Santos
Nov 17 at 21:32
add a comment |
Would $$lim_{xto0^+} |x|=lim_{xto0^-} |x|?$$ That is why I mentioned coming from either infinity to 0.
– kaisa
Nov 17 at 21:29
Both of those limits are equal to $0$. In other words, $lim_{xto0}lvert xrvert=0$.
– José Carlos Santos
Nov 17 at 21:32
Would $$lim_{xto0^+} |x|=lim_{xto0^-} |x|?$$ That is why I mentioned coming from either infinity to 0.
– kaisa
Nov 17 at 21:29
Would $$lim_{xto0^+} |x|=lim_{xto0^-} |x|?$$ That is why I mentioned coming from either infinity to 0.
– kaisa
Nov 17 at 21:29
Both of those limits are equal to $0$. In other words, $lim_{xto0}lvert xrvert=0$.
– José Carlos Santos
Nov 17 at 21:32
Both of those limits are equal to $0$. In other words, $lim_{xto0}lvert xrvert=0$.
– José Carlos Santos
Nov 17 at 21:32
add a comment |
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1
Why would you take the limit to $;pminfty;$ at all? The limit is when $;xto0;$ ...and that's all!
– DonAntonio
Nov 17 at 21:28