Real Analysis: Differentiability Proof [closed]
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 at 21:47
Martin Sleziak
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asked Nov 30 at 17:31
Nick
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closed as off-topic by José Carlos Santos, RRL, Paul Frost, Shailesh, John B Dec 1 at 0:44
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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 at 21:47
Martin Sleziak
44.6k7115270
asked Nov 30 at 17:31
Nick
344
closed as off-topic by José Carlos Santos, RRL, Paul Frost, Shailesh, John B Dec 1 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 |
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 at 21:47
Martin Sleziak
44.6k7115270
asked Nov 30 at 17:31
Nick
344
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 at 21:47
Martin Sleziak
44.6k7115270
asked Nov 30 at 17:31
Nick
344
edited Nov 30 at 21:47
Martin Sleziak
44.6k7115270
asked Nov 30 at 17:31
Nick
344
edited Nov 30 at 21:47
Martin Sleziak
44.6k7115270
edited Nov 30 at 21:47
Martin Sleziak
44.6k7115270
edited Nov 30 at 21:47
Martin Sleziak
44.6k7115270
44.6k7115270
asked Nov 30 at 17:31
Nick
344
asked Nov 30 at 17:31
Nick
344
asked Nov 30 at 17:31
Nick
344
344
closed as off-topic by José Carlos Santos, RRL, Paul Frost, Shailesh, John B Dec 1 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 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 |
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1 Answer
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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 at 18:15
StackTD
22k1947
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 18:15
StackTD
22k1947
add a comment |
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 at 18:15
StackTD
22k1947
add a comment |
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 at 18:15
StackTD
22k1947
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 at 18:15
StackTD
22k1947
edited Nov 30 at 18:23
answered Nov 30 at 18:15
StackTD
22k1947
answered Nov 30 at 18:15
StackTD
22k1947
answered Nov 30 at 18:15
StackTD
22k1947
22k1947
add a comment |
add a comment |
