Do I have this idea of antiderivatives correct?
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So is all it's saying that if there are two functions that have the same derivatives for every single $x$ in the interval, then $f(x) = g(x) + alpha$, means that the second function is just the exact same as $f(x)$ except raised or lowered?
calculus
asked Dec 11 '18 at 1:49
mingming
3856
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add a comment |
$begingroup$
So is all it's saying that if there are two functions that have the same derivatives for every single $x$ in the interval, then $f(x) = g(x) + alpha$, means that the second function is just the exact same as $f(x)$ except raised or lowered?
calculus
asked Dec 11 '18 at 1:49
mingming
3856
$endgroup$
1
$begingroup$
If by "raised or lowered", you mean that the graph of $g$ is the graph of $f$ translated up or down, yes, you seem to have a correct idea.
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– Xander Henderson
Dec 11 '18 at 1:52
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You should be able to prove your idea by a direct use of mean value theorem. Proving it without the use mean value theorem is hard and should convince you of the importance of mean value theorem.
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– Paramanand Singh
Dec 11 '18 at 6:17
add a comment |
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So is all it's saying that if there are two functions that have the same derivatives for every single $x$ in the interval, then $f(x) = g(x) + alpha$, means that the second function is just the exact same as $f(x)$ except raised or lowered?
calculus
asked Dec 11 '18 at 1:49
mingming
3856
$endgroup$
So is all it's saying that if there are two functions that have the same derivatives for every single $x$ in the interval, then $f(x) = g(x) + alpha$, means that the second function is just the exact same as $f(x)$ except raised or lowered?
calculus
calculus
asked Dec 11 '18 at 1:49
mingming
3856
asked Dec 11 '18 at 1:49
mingming
3856
asked Dec 11 '18 at 1:49
mingming
3856
asked Dec 11 '18 at 1:49
mingming
3856
asked Dec 11 '18 at 1:49
mingming
3856
3856
1
$begingroup$
If by "raised or lowered", you mean that the graph of $g$ is the graph of $f$ translated up or down, yes, you seem to have a correct idea.
$endgroup$
– Xander Henderson
Dec 11 '18 at 1:52
$begingroup$
You should be able to prove your idea by a direct use of mean value theorem. Proving it without the use mean value theorem is hard and should convince you of the importance of mean value theorem.
$endgroup$
– Paramanand Singh
Dec 11 '18 at 6:17
add a comment |
1
$begingroup$
If by "raised or lowered", you mean that the graph of $g$ is the graph of $f$ translated up or down, yes, you seem to have a correct idea.
$endgroup$
– Xander Henderson
Dec 11 '18 at 1:52
$begingroup$
You should be able to prove your idea by a direct use of mean value theorem. Proving it without the use mean value theorem is hard and should convince you of the importance of mean value theorem.
$endgroup$
– Paramanand Singh
Dec 11 '18 at 6:17
1
1
$begingroup$
If by "raised or lowered", you mean that the graph of $g$ is the graph of $f$ translated up or down, yes, you seem to have a correct idea.
$endgroup$
– Xander Henderson
Dec 11 '18 at 1:52
$begingroup$
If by "raised or lowered", you mean that the graph of $g$ is the graph of $f$ translated up or down, yes, you seem to have a correct idea.
$endgroup$
– Xander Henderson
Dec 11 '18 at 1:52
$begingroup$
You should be able to prove your idea by a direct use of mean value theorem. Proving it without the use mean value theorem is hard and should convince you of the importance of mean value theorem.
$endgroup$
– Paramanand Singh
Dec 11 '18 at 6:17
$begingroup$
You should be able to prove your idea by a direct use of mean value theorem. Proving it without the use mean value theorem is hard and should convince you of the importance of mean value theorem.
$endgroup$
– Paramanand Singh
Dec 11 '18 at 6:17
add a comment |
2 Answers
2
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oldest
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Yes, because the difference $f-g$ must then have zero derivative everywhere and therefore be constant. This means the graph of $f$ is the graph of $g$ shifted vertically.
answered Dec 11 '18 at 1:52
MPWMPW
30.3k12157
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add a comment |
$begingroup$
Yes, $int f(x) dx$ is a family of functions that only differ by a constant, which is typically denoted by $c$. This constant $c$ can be viewed as a shift vertically of the function.
The idea of a family of antiderivatives is easily shown by, say $f(x)=x^2+1$ and $g(x)=x^2-5$. While $f ne g$, it is true that $f'=g'$. Likewise $int 2x dx=x^2+c$, which contains $f$ and $g$.
answered Dec 11 '18 at 2:00
Dando18Dando18
4,68241235
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes, because the difference $f-g$ must then have zero derivative everywhere and therefore be constant. This means the graph of $f$ is the graph of $g$ shifted vertically.
answered Dec 11 '18 at 1:52
MPWMPW
30.3k12157
$endgroup$
add a comment |
$begingroup$
Yes, because the difference $f-g$ must then have zero derivative everywhere and therefore be constant. This means the graph of $f$ is the graph of $g$ shifted vertically.
answered Dec 11 '18 at 1:52
MPWMPW
30.3k12157
$endgroup$
add a comment |
$begingroup$
Yes, because the difference $f-g$ must then have zero derivative everywhere and therefore be constant. This means the graph of $f$ is the graph of $g$ shifted vertically.
answered Dec 11 '18 at 1:52
MPWMPW
30.3k12157
$endgroup$
Yes, because the difference $f-g$ must then have zero derivative everywhere and therefore be constant. This means the graph of $f$ is the graph of $g$ shifted vertically.
answered Dec 11 '18 at 1:52
MPWMPW
30.3k12157
answered Dec 11 '18 at 1:52
MPWMPW
30.3k12157
answered Dec 11 '18 at 1:52
MPWMPW
30.3k12157
answered Dec 11 '18 at 1:52
MPWMPW
30.3k12157
30.3k12157
add a comment |
add a comment |
$begingroup$
Yes, $int f(x) dx$ is a family of functions that only differ by a constant, which is typically denoted by $c$. This constant $c$ can be viewed as a shift vertically of the function.
The idea of a family of antiderivatives is easily shown by, say $f(x)=x^2+1$ and $g(x)=x^2-5$. While $f ne g$, it is true that $f'=g'$. Likewise $int 2x dx=x^2+c$, which contains $f$ and $g$.
answered Dec 11 '18 at 2:00
Dando18Dando18
4,68241235
$endgroup$
add a comment |
$begingroup$
Yes, $int f(x) dx$ is a family of functions that only differ by a constant, which is typically denoted by $c$. This constant $c$ can be viewed as a shift vertically of the function.
The idea of a family of antiderivatives is easily shown by, say $f(x)=x^2+1$ and $g(x)=x^2-5$. While $f ne g$, it is true that $f'=g'$. Likewise $int 2x dx=x^2+c$, which contains $f$ and $g$.
answered Dec 11 '18 at 2:00
Dando18Dando18
4,68241235
$endgroup$
add a comment |
$begingroup$
Yes, $int f(x) dx$ is a family of functions that only differ by a constant, which is typically denoted by $c$. This constant $c$ can be viewed as a shift vertically of the function.
The idea of a family of antiderivatives is easily shown by, say $f(x)=x^2+1$ and $g(x)=x^2-5$. While $f ne g$, it is true that $f'=g'$. Likewise $int 2x dx=x^2+c$, which contains $f$ and $g$.
answered Dec 11 '18 at 2:00
Dando18Dando18
4,68241235
$endgroup$
Yes, $int f(x) dx$ is a family of functions that only differ by a constant, which is typically denoted by $c$. This constant $c$ can be viewed as a shift vertically of the function.
The idea of a family of antiderivatives is easily shown by, say $f(x)=x^2+1$ and $g(x)=x^2-5$. While $f ne g$, it is true that $f'=g'$. Likewise $int 2x dx=x^2+c$, which contains $f$ and $g$.
answered Dec 11 '18 at 2:00
Dando18Dando18
4,68241235
answered Dec 11 '18 at 2:00
Dando18Dando18
4,68241235
answered Dec 11 '18 at 2:00
Dando18Dando18
4,68241235
answered Dec 11 '18 at 2:00
Dando18Dando18
4,68241235
4,68241235
add a comment |
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$begingroup$
If by "raised or lowered", you mean that the graph of $g$ is the graph of $f$ translated up or down, yes, you seem to have a correct idea.
$endgroup$
– Xander Henderson
Dec 11 '18 at 1:52
$begingroup$
You should be able to prove your idea by a direct use of mean value theorem. Proving it without the use mean value theorem is hard and should convince you of the importance of mean value theorem.
$endgroup$
– Paramanand Singh
Dec 11 '18 at 6:17