Real Analysis: Differentiability Proof [closed]
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I need help with the forward direction of this iff proof, I'm not really sure how to go about proving the existence of such a function...
real-analysis derivatives
edited Nov 30 '18 at 21:47
Martin Sleziak
44.7k8117272
asked Nov 30 '18 at 17:31
NickNick
344
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closed as off-topic by José Carlos Santos, RRL, Paul Frost, Shailesh, John B Dec 1 '18 at 0:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, RRL, Paul Frost, Shailesh, John B
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$begingroup$
I need help with the forward direction of this iff proof, I'm not really sure how to go about proving the existence of such a function...
real-analysis derivatives
edited Nov 30 '18 at 21:47
Martin Sleziak
44.7k8117272
asked Nov 30 '18 at 17:31
NickNick
344
$endgroup$
closed as off-topic by José Carlos Santos, RRL, Paul Frost, Shailesh, John B Dec 1 '18 at 0:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, RRL, Paul Frost, Shailesh, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I need help with the forward direction of this iff proof, I'm not really sure how to go about proving the existence of such a function...
real-analysis derivatives
edited Nov 30 '18 at 21:47
Martin Sleziak
44.7k8117272
asked Nov 30 '18 at 17:31
NickNick
344
$endgroup$
I need help with the forward direction of this iff proof, I'm not really sure how to go about proving the existence of such a function...
real-analysis derivatives
real-analysis derivatives
edited Nov 30 '18 at 21:47
Martin Sleziak
44.7k8117272
asked Nov 30 '18 at 17:31
NickNick
344
edited Nov 30 '18 at 21:47
Martin Sleziak
44.7k8117272
asked Nov 30 '18 at 17:31
NickNick
344
edited Nov 30 '18 at 21:47
Martin Sleziak
44.7k8117272
edited Nov 30 '18 at 21:47
Martin Sleziak
44.7k8117272
edited Nov 30 '18 at 21:47
Martin Sleziak
44.7k8117272
44.7k8117272
asked Nov 30 '18 at 17:31
NickNick
344
asked Nov 30 '18 at 17:31
NickNick
344
asked Nov 30 '18 at 17:31
NickNick
344
344
closed as off-topic by José Carlos Santos, RRL, Paul Frost, Shailesh, John B Dec 1 '18 at 0:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, RRL, Paul Frost, Shailesh, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, RRL, Paul Frost, Shailesh, John B Dec 1 '18 at 0:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, RRL, Paul Frost, Shailesh, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
Notice that $lambda$ will exactly be the derivative of $f$ in $x_0$ and use this, together with the expression you're given, to get an idea of how to define $epsilon(x)$; i.e. simply solve for $epsilon(x)$ and see if that works.
Suppose $f$ is differentiable at $x_0$ and call its derivative $f'(x_0)$. Now define:
$$epsilon(x)=f(x)-f(x_0)-f'(x_0)(x-x_0)$$
You then have:
$$frac{epsilon(x)}{x-x_0}=frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)$$
And so:
$$lim_{xto x_0}frac{epsilon(x)}{x-x_0}=lim_{xto x_0}left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right) = ldots$$
answered Nov 30 '18 at 18:15
StackTDStackTD
22.5k2049
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that $lambda$ will exactly be the derivative of $f$ in $x_0$ and use this, together with the expression you're given, to get an idea of how to define $epsilon(x)$; i.e. simply solve for $epsilon(x)$ and see if that works.
Suppose $f$ is differentiable at $x_0$ and call its derivative $f'(x_0)$. Now define:
$$epsilon(x)=f(x)-f(x_0)-f'(x_0)(x-x_0)$$
You then have:
$$frac{epsilon(x)}{x-x_0}=frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)$$
And so:
$$lim_{xto x_0}frac{epsilon(x)}{x-x_0}=lim_{xto x_0}left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right) = ldots$$
answered Nov 30 '18 at 18:15
StackTDStackTD
22.5k2049
$endgroup$
add a comment |
$begingroup$
Notice that $lambda$ will exactly be the derivative of $f$ in $x_0$ and use this, together with the expression you're given, to get an idea of how to define $epsilon(x)$; i.e. simply solve for $epsilon(x)$ and see if that works.
Suppose $f$ is differentiable at $x_0$ and call its derivative $f'(x_0)$. Now define:
$$epsilon(x)=f(x)-f(x_0)-f'(x_0)(x-x_0)$$
You then have:
$$frac{epsilon(x)}{x-x_0}=frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)$$
And so:
$$lim_{xto x_0}frac{epsilon(x)}{x-x_0}=lim_{xto x_0}left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right) = ldots$$
answered Nov 30 '18 at 18:15
StackTDStackTD
22.5k2049
$endgroup$
add a comment |
$begingroup$
Notice that $lambda$ will exactly be the derivative of $f$ in $x_0$ and use this, together with the expression you're given, to get an idea of how to define $epsilon(x)$; i.e. simply solve for $epsilon(x)$ and see if that works.
Suppose $f$ is differentiable at $x_0$ and call its derivative $f'(x_0)$. Now define:
$$epsilon(x)=f(x)-f(x_0)-f'(x_0)(x-x_0)$$
You then have:
$$frac{epsilon(x)}{x-x_0}=frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)$$
And so:
$$lim_{xto x_0}frac{epsilon(x)}{x-x_0}=lim_{xto x_0}left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right) = ldots$$
answered Nov 30 '18 at 18:15
StackTDStackTD
22.5k2049
$endgroup$
Notice that $lambda$ will exactly be the derivative of $f$ in $x_0$ and use this, together with the expression you're given, to get an idea of how to define $epsilon(x)$; i.e. simply solve for $epsilon(x)$ and see if that works.
Suppose $f$ is differentiable at $x_0$ and call its derivative $f'(x_0)$. Now define:
$$epsilon(x)=f(x)-f(x_0)-f'(x_0)(x-x_0)$$
You then have:
$$frac{epsilon(x)}{x-x_0}=frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)$$
And so:
$$lim_{xto x_0}frac{epsilon(x)}{x-x_0}=lim_{xto x_0}left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right) = ldots$$
answered Nov 30 '18 at 18:15
StackTDStackTD
22.5k2049
edited Nov 30 '18 at 18:23
answered Nov 30 '18 at 18:15
StackTDStackTD
22.5k2049
answered Nov 30 '18 at 18:15
StackTDStackTD
22.5k2049
answered Nov 30 '18 at 18:15
StackTDStackTD
22.5k2049
22.5k2049
add a comment |
add a comment |
