Mixing
Integrating functions in real analysis
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The following is a True/False problem from an exam I got back:
If $f'(x)=1$ for each $xinmathbbR$ and $f(2)=7$ then $f(-2)=-7$.
I said false for this problem, but it was marked wrong. My reasoning for this was to treat this as an initial value problem. First, integrate $f'(x)$ so that $f(x)=int f'(x)dx=x+c$. Next solve for $c$ from $f(2)=7$. Thus $2+c=7implies c=5$. So our function is $f(x)=x+5$ where $f'(x)=1forall xinmathbbR$ and $f(2)=7$. Both of the statements in the AND statement in the hypothesis are true, therefore the hypothesis is true. Next, checking $f(-2)=3neq-7$ therefore the conclusion of the implication is false. Here we have True imply false, therefore the implication is false.
Is there an error in my rationale?
real-analysis proof-verification
asked Jul 23 at 22:30
Peetrius
358111
add a comment |Â
up vote
1
down vote
favorite
The following is a True/False problem from an exam I got back:
If $f'(x)=1$ for each $xinmathbbR$ and $f(2)=7$ then $f(-2)=-7$.
I said false for this problem, but it was marked wrong. My reasoning for this was to treat this as an initial value problem. First, integrate $f'(x)$ so that $f(x)=int f'(x)dx=x+c$. Next solve for $c$ from $f(2)=7$. Thus $2+c=7implies c=5$. So our function is $f(x)=x+5$ where $f'(x)=1forall xinmathbbR$ and $f(2)=7$. Both of the statements in the AND statement in the hypothesis are true, therefore the hypothesis is true. Next, checking $f(-2)=3neq-7$ therefore the conclusion of the implication is false. Here we have True imply false, therefore the implication is false.
Is there an error in my rationale?
real-analysis proof-verification
asked Jul 23 at 22:30
Peetrius
358111
1
Everything you show is correct. My best thought is if you have not discussed integration within the context of the real analysis class, you must use only the theorems / definitions covered thus far in class, and need to use an argument based on the mean value theorem or similar.
– Doug M
Jul 23 at 22:36
The exam was graded on a raw score of whether or not we chose the correct answer, so no explanations were necessary. In the post, I wanted to be confident in my response so I wrote out my thoughts. Also, I didn't want to appear that I wasn't merely fishing for answers. Though you do have a point with the explanation here. This problem was testing on the MVT, not my skills to solve IVP.
– Peetrius
Jul 23 at 22:39
One potential problem is that what if the ODE has a non-unique solution (well it does as it's a first order linear ODE with CC by a famous theorem). But other than that, it looks correct.
– Raymond Chu
Jul 24 at 0:50
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The following is a True/False problem from an exam I got back:
If $f'(x)=1$ for each $xinmathbbR$ and $f(2)=7$ then $f(-2)=-7$.
I said false for this problem, but it was marked wrong. My reasoning for this was to treat this as an initial value problem. First, integrate $f'(x)$ so that $f(x)=int f'(x)dx=x+c$. Next solve for $c$ from $f(2)=7$. Thus $2+c=7implies c=5$. So our function is $f(x)=x+5$ where $f'(x)=1forall xinmathbbR$ and $f(2)=7$. Both of the statements in the AND statement in the hypothesis are true, therefore the hypothesis is true. Next, checking $f(-2)=3neq-7$ therefore the conclusion of the implication is false. Here we have True imply false, therefore the implication is false.
Is there an error in my rationale?
real-analysis proof-verification
asked Jul 23 at 22:30
Peetrius
358111
The following is a True/False problem from an exam I got back:
If $f'(x)=1$ for each $xinmathbbR$ and $f(2)=7$ then $f(-2)=-7$.
I said false for this problem, but it was marked wrong. My reasoning for this was to treat this as an initial value problem. First, integrate $f'(x)$ so that $f(x)=int f'(x)dx=x+c$. Next solve for $c$ from $f(2)=7$. Thus $2+c=7implies c=5$. So our function is $f(x)=x+5$ where $f'(x)=1forall xinmathbbR$ and $f(2)=7$. Both of the statements in the AND statement in the hypothesis are true, therefore the hypothesis is true. Next, checking $f(-2)=3neq-7$ therefore the conclusion of the implication is false. Here we have True imply false, therefore the implication is false.
Is there an error in my rationale?
real-analysis proof-verification
asked Jul 23 at 22:30
Peetrius
358111
asked Jul 23 at 22:30
Peetrius
358111
asked Jul 23 at 22:30
Peetrius
358111
asked Jul 23 at 22:30
Peetrius
358111
358111
1
Everything you show is correct. My best thought is if you have not discussed integration within the context of the real analysis class, you must use only the theorems / definitions covered thus far in class, and need to use an argument based on the mean value theorem or similar.
– Doug M
Jul 23 at 22:36
The exam was graded on a raw score of whether or not we chose the correct answer, so no explanations were necessary. In the post, I wanted to be confident in my response so I wrote out my thoughts. Also, I didn't want to appear that I wasn't merely fishing for answers. Though you do have a point with the explanation here. This problem was testing on the MVT, not my skills to solve IVP.
– Peetrius
Jul 23 at 22:39
One potential problem is that what if the ODE has a non-unique solution (well it does as it's a first order linear ODE with CC by a famous theorem). But other than that, it looks correct.
– Raymond Chu
Jul 24 at 0:50
add a comment |Â
1
Everything you show is correct. My best thought is if you have not discussed integration within the context of the real analysis class, you must use only the theorems / definitions covered thus far in class, and need to use an argument based on the mean value theorem or similar.
– Doug M
Jul 23 at 22:36
The exam was graded on a raw score of whether or not we chose the correct answer, so no explanations were necessary. In the post, I wanted to be confident in my response so I wrote out my thoughts. Also, I didn't want to appear that I wasn't merely fishing for answers. Though you do have a point with the explanation here. This problem was testing on the MVT, not my skills to solve IVP.
– Peetrius
Jul 23 at 22:39
One potential problem is that what if the ODE has a non-unique solution (well it does as it's a first order linear ODE with CC by a famous theorem). But other than that, it looks correct.
– Raymond Chu
Jul 24 at 0:50
1
1
Everything you show is correct. My best thought is if you have not discussed integration within the context of the real analysis class, you must use only the theorems / definitions covered thus far in class, and need to use an argument based on the mean value theorem or similar.
– Doug M
Jul 23 at 22:36
Everything you show is correct. My best thought is if you have not discussed integration within the context of the real analysis class, you must use only the theorems / definitions covered thus far in class, and need to use an argument based on the mean value theorem or similar.
– Doug M
Jul 23 at 22:36
The exam was graded on a raw score of whether or not we chose the correct answer, so no explanations were necessary. In the post, I wanted to be confident in my response so I wrote out my thoughts. Also, I didn't want to appear that I wasn't merely fishing for answers. Though you do have a point with the explanation here. This problem was testing on the MVT, not my skills to solve IVP.
– Peetrius
Jul 23 at 22:39
The exam was graded on a raw score of whether or not we chose the correct answer, so no explanations were necessary. In the post, I wanted to be confident in my response so I wrote out my thoughts. Also, I didn't want to appear that I wasn't merely fishing for answers. Though you do have a point with the explanation here. This problem was testing on the MVT, not my skills to solve IVP.
– Peetrius
Jul 23 at 22:39
One potential problem is that what if the ODE has a non-unique solution (well it does as it's a first order linear ODE with CC by a famous theorem). But other than that, it looks correct.
– Raymond Chu
Jul 24 at 0:50
One potential problem is that what if the ODE has a non-unique solution (well it does as it's a first order linear ODE with CC by a famous theorem). But other than that, it looks correct.
– Raymond Chu
Jul 24 at 0:50
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Your argument is correct. Besides, note that$$fracf(2)-f(-2)2-(-2)=frac144=frac72.$$But, by the mean value theorem, that quotient should be $1$.
answered Jul 23 at 22:34
José Carlos Santos
113k1698176
A classmate noted that she used the MVT for this problem. I didn't remember how exactly she used it, but reading this I can see that your argument matched hers. Much simpler explanation. Thanks!
– Peetrius
Jul 23 at 22:36
1
I'm glad I could help.
– José Carlos Santos
Jul 23 at 22:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Your argument is correct. Besides, note that$$fracf(2)-f(-2)2-(-2)=frac144=frac72.$$But, by the mean value theorem, that quotient should be $1$.
answered Jul 23 at 22:34
José Carlos Santos
113k1698176
A classmate noted that she used the MVT for this problem. I didn't remember how exactly she used it, but reading this I can see that your argument matched hers. Much simpler explanation. Thanks!
– Peetrius
Jul 23 at 22:36
1
I'm glad I could help.
– José Carlos Santos
Jul 23 at 22:48
add a comment |Â
up vote
5
down vote
accepted
Your argument is correct. Besides, note that$$fracf(2)-f(-2)2-(-2)=frac144=frac72.$$But, by the mean value theorem, that quotient should be $1$.
answered Jul 23 at 22:34
José Carlos Santos
113k1698176
A classmate noted that she used the MVT for this problem. I didn't remember how exactly she used it, but reading this I can see that your argument matched hers. Much simpler explanation. Thanks!
– Peetrius
Jul 23 at 22:36
1
I'm glad I could help.
– José Carlos Santos
Jul 23 at 22:48
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Your argument is correct. Besides, note that$$fracf(2)-f(-2)2-(-2)=frac144=frac72.$$But, by the mean value theorem, that quotient should be $1$.
answered Jul 23 at 22:34
José Carlos Santos
113k1698176
Your argument is correct. Besides, note that$$fracf(2)-f(-2)2-(-2)=frac144=frac72.$$But, by the mean value theorem, that quotient should be $1$.
answered Jul 23 at 22:34
José Carlos Santos
113k1698176
answered Jul 23 at 22:34
José Carlos Santos
113k1698176
answered Jul 23 at 22:34
José Carlos Santos
113k1698176
answered Jul 23 at 22:34
José Carlos Santos
113k1698176
113k1698176
A classmate noted that she used the MVT for this problem. I didn't remember how exactly she used it, but reading this I can see that your argument matched hers. Much simpler explanation. Thanks!
– Peetrius
Jul 23 at 22:36
1
I'm glad I could help.
– José Carlos Santos
Jul 23 at 22:48
add a comment |Â
A classmate noted that she used the MVT for this problem. I didn't remember how exactly she used it, but reading this I can see that your argument matched hers. Much simpler explanation. Thanks!
– Peetrius
Jul 23 at 22:36
1
I'm glad I could help.
– José Carlos Santos
Jul 23 at 22:48
A classmate noted that she used the MVT for this problem. I didn't remember how exactly she used it, but reading this I can see that your argument matched hers. Much simpler explanation. Thanks!
– Peetrius
Jul 23 at 22:36
A classmate noted that she used the MVT for this problem. I didn't remember how exactly she used it, but reading this I can see that your argument matched hers. Much simpler explanation. Thanks!
– Peetrius
Jul 23 at 22:36
1
1
I'm glad I could help.
– José Carlos Santos
Jul 23 at 22:48
I'm glad I could help.
– José Carlos Santos
Jul 23 at 22:48
add a comment |Â
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1
Everything you show is correct. My best thought is if you have not discussed integration within the context of the real analysis class, you must use only the theorems / definitions covered thus far in class, and need to use an argument based on the mean value theorem or similar.
– Doug M
Jul 23 at 22:36
The exam was graded on a raw score of whether or not we chose the correct answer, so no explanations were necessary. In the post, I wanted to be confident in my response so I wrote out my thoughts. Also, I didn't want to appear that I wasn't merely fishing for answers. Though you do have a point with the explanation here. This problem was testing on the MVT, not my skills to solve IVP.
– Peetrius
Jul 23 at 22:39
One potential problem is that what if the ODE has a non-unique solution (well it does as it's a first order linear ODE with CC by a famous theorem). But other than that, it looks correct.
– Raymond Chu
Jul 24 at 0:50