Differentiable proofs Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)An example of differentiability in $mathbbR^n$ everywhere but not at origin.Uniformly differentiable functionShowing a piece-wise function is differentiable everywhere/Clarification$f$ is differentiable twice but not in $0$?differentiable function proofsTrying to Understand How to write ProofsIf $f$ is differentiable on $[1,2]$, then $exists alphain(1,2): f(2)-f(1) = fracalpha^22f'(alpha)$Let $f:(0,1] rightarrow mathbbR$ be differentiable on $(0,1]$, with $|f'(x)| leq 1$ let $a_n=f(1/n)$ Show that $(a_n)$converges.existence of a certain differentiable function$C^1$ Parameter Transformation. Show function and it's inverse are continuously differentiable.
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Differentiable proofs
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)An example of differentiability in $mathbbR^n$ everywhere but not at origin.Uniformly differentiable functionShowing a piece-wise function is differentiable everywhere/Clarification$f$ is differentiable twice but not in $0$?differentiable function proofsTrying to Understand How to write ProofsIf $f$ is differentiable on $[1,2]$, then $exists alphain(1,2): f(2)-f(1) = fracalpha^22f'(alpha)$Let $f:(0,1] rightarrow mathbbR$ be differentiable on $(0,1]$, with $|f'(x)| leq 1$ let $a_n=f(1/n)$ Show that $(a_n)$converges.existence of a certain differentiable function$C^1$ Parameter Transformation. Show function and it's inverse are continuously differentiable.
$begingroup$
Show that if $f:mathbb Rto mathbb R$ is such that $-x^2le f(x)le x^2$ for all $xin mathbb R$, then $f$ is differentiable at $x=0$ and $f'(0)=0$.
Really confused because isnt 0 the only soltion. and How would I apply the definition of differentiable to this?
calculus proof-writing
asked Apr 1 at 17:36
DoubleliftDoublelift
305
$endgroup$
add a comment |
$begingroup$
Show that if $f:mathbb Rto mathbb R$ is such that $-x^2le f(x)le x^2$ for all $xin mathbb R$, then $f$ is differentiable at $x=0$ and $f'(0)=0$.
Really confused because isnt 0 the only soltion. and How would I apply the definition of differentiable to this?
calculus proof-writing
asked Apr 1 at 17:36
DoubleliftDoublelift
305
$endgroup$
$begingroup$
What do you mean "0 is the only solution"? What the inequality says is just that the curve $y=f(x)$ lies between the curves $y=x^2$ and $y=-x^2$, and there are many such functions.
$endgroup$
– Hans Lundmark
Apr 1 at 19:36
$begingroup$
My bad I did not really get what was being asked
$endgroup$
– Doublelift
Apr 1 at 19:37
add a comment |
$begingroup$
Show that if $f:mathbb Rto mathbb R$ is such that $-x^2le f(x)le x^2$ for all $xin mathbb R$, then $f$ is differentiable at $x=0$ and $f'(0)=0$.
Really confused because isnt 0 the only soltion. and How would I apply the definition of differentiable to this?
calculus proof-writing
asked Apr 1 at 17:36
DoubleliftDoublelift
305
$endgroup$
Show that if $f:mathbb Rto mathbb R$ is such that $-x^2le f(x)le x^2$ for all $xin mathbb R$, then $f$ is differentiable at $x=0$ and $f'(0)=0$.
Really confused because isnt 0 the only soltion. and How would I apply the definition of differentiable to this?
calculus proof-writing
calculus proof-writing
asked Apr 1 at 17:36
DoubleliftDoublelift
305
asked Apr 1 at 17:36
DoubleliftDoublelift
305
asked Apr 1 at 17:36
DoubleliftDoublelift
305
asked Apr 1 at 17:36
DoubleliftDoublelift
305
asked Apr 1 at 17:36
DoubleliftDoublelift
305
305
$begingroup$
What do you mean "0 is the only solution"? What the inequality says is just that the curve $y=f(x)$ lies between the curves $y=x^2$ and $y=-x^2$, and there are many such functions.
$endgroup$
– Hans Lundmark
Apr 1 at 19:36
$begingroup$
My bad I did not really get what was being asked
$endgroup$
– Doublelift
Apr 1 at 19:37
add a comment |
$begingroup$
What do you mean "0 is the only solution"? What the inequality says is just that the curve $y=f(x)$ lies between the curves $y=x^2$ and $y=-x^2$, and there are many such functions.
$endgroup$
– Hans Lundmark
Apr 1 at 19:36
$begingroup$
My bad I did not really get what was being asked
$endgroup$
– Doublelift
Apr 1 at 19:37
$begingroup$
What do you mean "0 is the only solution"? What the inequality says is just that the curve $y=f(x)$ lies between the curves $y=x^2$ and $y=-x^2$, and there are many such functions.
$endgroup$
– Hans Lundmark
Apr 1 at 19:36
$begingroup$
What do you mean "0 is the only solution"? What the inequality says is just that the curve $y=f(x)$ lies between the curves $y=x^2$ and $y=-x^2$, and there are many such functions.
$endgroup$
– Hans Lundmark
Apr 1 at 19:36
$begingroup$
My bad I did not really get what was being asked
$endgroup$
– Doublelift
Apr 1 at 19:37
$begingroup$
My bad I did not really get what was being asked
$endgroup$
– Doublelift
Apr 1 at 19:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $h neq 0$ be small and note that the given inequality implies that $f(0) = 0$ (take $x=0$ in the inequality). Therefore,
beginalign*
fracf(0+h)-f(0)h = fracf(h)h.
endalign*
Since $-x^2 leq f(x) leq x^2$ implies $|f(x)| leq x^2$, it follows that
beginalign*
leftvertfracf(0+h)-f(0)hrightvert = leftvertfracf(h)h rightvert
leq frach^2 = |h| to 0
endalign*
as $h to 0$. By definition of the derivative, we see that $f^prime(0)$ exists and is equal to $0$.
answered Apr 1 at 17:39
rolandcyprolandcyp
2,149422
$endgroup$
$begingroup$
So being greater then -x^2 wont have a effect on the h you output?
$endgroup$
– Doublelift
Apr 1 at 17:48
$begingroup$
@JJBurt Having that $-x^2 leq f(x) leq x^2$ tells us that $|f(x)| leq x^2$. This is exactly what is needed for the argument
$endgroup$
– rolandcyp
Apr 1 at 17:49
$begingroup$
@JJBurt When working with limits, it is generally much more helpful to have both an upper and a lower bound for the function!
$endgroup$
– rolandcyp
Apr 1 at 17:51
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $h neq 0$ be small and note that the given inequality implies that $f(0) = 0$ (take $x=0$ in the inequality). Therefore,
beginalign*
fracf(0+h)-f(0)h = fracf(h)h.
endalign*
Since $-x^2 leq f(x) leq x^2$ implies $|f(x)| leq x^2$, it follows that
beginalign*
leftvertfracf(0+h)-f(0)hrightvert = leftvertfracf(h)h rightvert
leq frach^2 = |h| to 0
endalign*
as $h to 0$. By definition of the derivative, we see that $f^prime(0)$ exists and is equal to $0$.
answered Apr 1 at 17:39
rolandcyprolandcyp
2,149422
$endgroup$
$begingroup$
So being greater then -x^2 wont have a effect on the h you output?
$endgroup$
– Doublelift
Apr 1 at 17:48
$begingroup$
@JJBurt Having that $-x^2 leq f(x) leq x^2$ tells us that $|f(x)| leq x^2$. This is exactly what is needed for the argument
$endgroup$
– rolandcyp
Apr 1 at 17:49
$begingroup$
@JJBurt When working with limits, it is generally much more helpful to have both an upper and a lower bound for the function!
$endgroup$
– rolandcyp
Apr 1 at 17:51
add a comment |
$begingroup$
Let $h neq 0$ be small and note that the given inequality implies that $f(0) = 0$ (take $x=0$ in the inequality). Therefore,
beginalign*
fracf(0+h)-f(0)h = fracf(h)h.
endalign*
Since $-x^2 leq f(x) leq x^2$ implies $|f(x)| leq x^2$, it follows that
beginalign*
leftvertfracf(0+h)-f(0)hrightvert = leftvertfracf(h)h rightvert
leq frach^2 = |h| to 0
endalign*
as $h to 0$. By definition of the derivative, we see that $f^prime(0)$ exists and is equal to $0$.
answered Apr 1 at 17:39
rolandcyprolandcyp
2,149422
$endgroup$
$begingroup$
So being greater then -x^2 wont have a effect on the h you output?
$endgroup$
– Doublelift
Apr 1 at 17:48
$begingroup$
@JJBurt Having that $-x^2 leq f(x) leq x^2$ tells us that $|f(x)| leq x^2$. This is exactly what is needed for the argument
$endgroup$
– rolandcyp
Apr 1 at 17:49
$begingroup$
@JJBurt When working with limits, it is generally much more helpful to have both an upper and a lower bound for the function!
$endgroup$
– rolandcyp
Apr 1 at 17:51
add a comment |
$begingroup$
Let $h neq 0$ be small and note that the given inequality implies that $f(0) = 0$ (take $x=0$ in the inequality). Therefore,
beginalign*
fracf(0+h)-f(0)h = fracf(h)h.
endalign*
Since $-x^2 leq f(x) leq x^2$ implies $|f(x)| leq x^2$, it follows that
beginalign*
leftvertfracf(0+h)-f(0)hrightvert = leftvertfracf(h)h rightvert
leq frach^2 = |h| to 0
endalign*
as $h to 0$. By definition of the derivative, we see that $f^prime(0)$ exists and is equal to $0$.
answered Apr 1 at 17:39
rolandcyprolandcyp
2,149422
$endgroup$
Let $h neq 0$ be small and note that the given inequality implies that $f(0) = 0$ (take $x=0$ in the inequality). Therefore,
beginalign*
fracf(0+h)-f(0)h = fracf(h)h.
endalign*
Since $-x^2 leq f(x) leq x^2$ implies $|f(x)| leq x^2$, it follows that
beginalign*
leftvertfracf(0+h)-f(0)hrightvert = leftvertfracf(h)h rightvert
leq frach^2 = |h| to 0
endalign*
as $h to 0$. By definition of the derivative, we see that $f^prime(0)$ exists and is equal to $0$.
answered Apr 1 at 17:39
rolandcyprolandcyp
2,149422
answered Apr 1 at 17:39
rolandcyprolandcyp
2,149422
answered Apr 1 at 17:39
rolandcyprolandcyp
2,149422
answered Apr 1 at 17:39
rolandcyprolandcyp
2,149422
2,149422
$begingroup$
So being greater then -x^2 wont have a effect on the h you output?
$endgroup$
– Doublelift
Apr 1 at 17:48
$begingroup$
@JJBurt Having that $-x^2 leq f(x) leq x^2$ tells us that $|f(x)| leq x^2$. This is exactly what is needed for the argument
$endgroup$
– rolandcyp
Apr 1 at 17:49
$begingroup$
@JJBurt When working with limits, it is generally much more helpful to have both an upper and a lower bound for the function!
$endgroup$
– rolandcyp
Apr 1 at 17:51
add a comment |
$begingroup$
So being greater then -x^2 wont have a effect on the h you output?
$endgroup$
– Doublelift
Apr 1 at 17:48
$begingroup$
@JJBurt Having that $-x^2 leq f(x) leq x^2$ tells us that $|f(x)| leq x^2$. This is exactly what is needed for the argument
$endgroup$
– rolandcyp
Apr 1 at 17:49
$begingroup$
@JJBurt When working with limits, it is generally much more helpful to have both an upper and a lower bound for the function!
$endgroup$
– rolandcyp
Apr 1 at 17:51
$begingroup$
So being greater then -x^2 wont have a effect on the h you output?
$endgroup$
– Doublelift
Apr 1 at 17:48
$begingroup$
So being greater then -x^2 wont have a effect on the h you output?
$endgroup$
– Doublelift
Apr 1 at 17:48
$begingroup$
@JJBurt Having that $-x^2 leq f(x) leq x^2$ tells us that $|f(x)| leq x^2$. This is exactly what is needed for the argument
$endgroup$
– rolandcyp
Apr 1 at 17:49
$begingroup$
@JJBurt Having that $-x^2 leq f(x) leq x^2$ tells us that $|f(x)| leq x^2$. This is exactly what is needed for the argument
$endgroup$
– rolandcyp
Apr 1 at 17:49
$begingroup$
@JJBurt When working with limits, it is generally much more helpful to have both an upper and a lower bound for the function!
$endgroup$
– rolandcyp
Apr 1 at 17:51
$begingroup$
@JJBurt When working with limits, it is generally much more helpful to have both an upper and a lower bound for the function!
$endgroup$
– rolandcyp
Apr 1 at 17:51
add a comment |
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$begingroup$
What do you mean "0 is the only solution"? What the inequality says is just that the curve $y=f(x)$ lies between the curves $y=x^2$ and $y=-x^2$, and there are many such functions.
$endgroup$
– Hans Lundmark
Apr 1 at 19:36
$begingroup$
My bad I did not really get what was being asked
$endgroup$
– Doublelift
Apr 1 at 19:37