if $f$ is continuous and differentiable on $(a,b)$ and $f'$ is positive for all but finitely many points then...
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Here's what my textbook asks me to prove:
Suppose that $f$ is continuous on $(a,b)$ and that $f'$ exists and is
positive for all but, finite number of points in $(a,b)$. Prove that
$f$ is strictly increasing on $(a,b)$.
My attempt:
Let $x_1 , x_2 , ldots x_n$ be points in $(a,b)$ where $f'(x_i)le 0$. We eliminate the possibility that $f'(x_i)< 0$ for any $i$.
Since $f'(x_i) < 0$, there exist $delta > 0$ , such that $frac{f(x)-f(x_i)}{x-x_i} < f'(x_i)/2 < 0$ for all $0<|x-x_i| < delta $ (This follows immediately from the definition of the limit). That is, $f(x) < 0$ for all $0<|x-x_i| < delta $ which contradicts our assumption that f is not positive at finitely many points. Thus it must be that $f'(x_i) = 0$ for all $i$.
Without any loss of generality, assume that $x_1 < x_2 < ldots < x_n $. Since f is continuous on $[x_i , x_{i+1}]$ and is differentiable on $(x_i , x_{i+1})$ and $f'(x) > 0$ on $(x_i , x_{i+1})$, we have that $f$ is strictly increasing on $[x_i , x_{i+1}]$. Thus, $f$ is strictly increasing on $[x_1 , x_n ]$.
I'm stuck at the part $(a, x_1]$ and $[x_n , b)$. I cannot apply the MVT here because $f$ is not continuous at $a$ and $b$. Also, is my rest of the proof correct?
real-analysis proof-verification
asked Nov 24 at 5:55
Ashish K
776513
add a comment |
up vote
1
down vote
favorite
Here's what my textbook asks me to prove:
Suppose that $f$ is continuous on $(a,b)$ and that $f'$ exists and is
positive for all but, finite number of points in $(a,b)$. Prove that
$f$ is strictly increasing on $(a,b)$.
My attempt:
Let $x_1 , x_2 , ldots x_n$ be points in $(a,b)$ where $f'(x_i)le 0$. We eliminate the possibility that $f'(x_i)< 0$ for any $i$.
Since $f'(x_i) < 0$, there exist $delta > 0$ , such that $frac{f(x)-f(x_i)}{x-x_i} < f'(x_i)/2 < 0$ for all $0<|x-x_i| < delta $ (This follows immediately from the definition of the limit). That is, $f(x) < 0$ for all $0<|x-x_i| < delta $ which contradicts our assumption that f is not positive at finitely many points. Thus it must be that $f'(x_i) = 0$ for all $i$.
Without any loss of generality, assume that $x_1 < x_2 < ldots < x_n $. Since f is continuous on $[x_i , x_{i+1}]$ and is differentiable on $(x_i , x_{i+1})$ and $f'(x) > 0$ on $(x_i , x_{i+1})$, we have that $f$ is strictly increasing on $[x_i , x_{i+1}]$. Thus, $f$ is strictly increasing on $[x_1 , x_n ]$.
I'm stuck at the part $(a, x_1]$ and $[x_n , b)$. I cannot apply the MVT here because $f$ is not continuous at $a$ and $b$. Also, is my rest of the proof correct?
real-analysis proof-verification
asked Nov 24 at 5:55
Ashish K
776513
1
Your proof is fine.The mean value theorem shows that $f$ is strictly increasing on all of the intervals $(a,x_1]$, $[x_k,x_{k+1}]$ and $[x_n,b)$ separately. It is straightforward to show that if $f$ is strictly increasing on $[p_1,p_2]$ and $[p_2,p_3]$ separately then $f$ is strictly increasing on $[p_1,p_3]$.
– copper.hat
Nov 24 at 6:24
@copper.hat it seems I've done a mistake. I've proved that $f(x)<f(x_i)$ but concluded the wrong thing. I need to fix it.
– Ashish K
Nov 24 at 7:04
I don't think you need the first paragraph of your attempt. I'm not sure where you think your mistake is. The proof is not complete, but you have the right idea.
– copper.hat
Nov 24 at 7:32
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here's what my textbook asks me to prove:
Suppose that $f$ is continuous on $(a,b)$ and that $f'$ exists and is
positive for all but, finite number of points in $(a,b)$. Prove that
$f$ is strictly increasing on $(a,b)$.
My attempt:
Let $x_1 , x_2 , ldots x_n$ be points in $(a,b)$ where $f'(x_i)le 0$. We eliminate the possibility that $f'(x_i)< 0$ for any $i$.
Since $f'(x_i) < 0$, there exist $delta > 0$ , such that $frac{f(x)-f(x_i)}{x-x_i} < f'(x_i)/2 < 0$ for all $0<|x-x_i| < delta $ (This follows immediately from the definition of the limit). That is, $f(x) < 0$ for all $0<|x-x_i| < delta $ which contradicts our assumption that f is not positive at finitely many points. Thus it must be that $f'(x_i) = 0$ for all $i$.
Without any loss of generality, assume that $x_1 < x_2 < ldots < x_n $. Since f is continuous on $[x_i , x_{i+1}]$ and is differentiable on $(x_i , x_{i+1})$ and $f'(x) > 0$ on $(x_i , x_{i+1})$, we have that $f$ is strictly increasing on $[x_i , x_{i+1}]$. Thus, $f$ is strictly increasing on $[x_1 , x_n ]$.
I'm stuck at the part $(a, x_1]$ and $[x_n , b)$. I cannot apply the MVT here because $f$ is not continuous at $a$ and $b$. Also, is my rest of the proof correct?
real-analysis proof-verification
asked Nov 24 at 5:55
Ashish K
776513
Here's what my textbook asks me to prove:
Suppose that $f$ is continuous on $(a,b)$ and that $f'$ exists and is
positive for all but, finite number of points in $(a,b)$. Prove that
$f$ is strictly increasing on $(a,b)$.
My attempt:
Let $x_1 , x_2 , ldots x_n$ be points in $(a,b)$ where $f'(x_i)le 0$. We eliminate the possibility that $f'(x_i)< 0$ for any $i$.
Since $f'(x_i) < 0$, there exist $delta > 0$ , such that $frac{f(x)-f(x_i)}{x-x_i} < f'(x_i)/2 < 0$ for all $0<|x-x_i| < delta $ (This follows immediately from the definition of the limit). That is, $f(x) < 0$ for all $0<|x-x_i| < delta $ which contradicts our assumption that f is not positive at finitely many points. Thus it must be that $f'(x_i) = 0$ for all $i$.
Without any loss of generality, assume that $x_1 < x_2 < ldots < x_n $. Since f is continuous on $[x_i , x_{i+1}]$ and is differentiable on $(x_i , x_{i+1})$ and $f'(x) > 0$ on $(x_i , x_{i+1})$, we have that $f$ is strictly increasing on $[x_i , x_{i+1}]$. Thus, $f$ is strictly increasing on $[x_1 , x_n ]$.
I'm stuck at the part $(a, x_1]$ and $[x_n , b)$. I cannot apply the MVT here because $f$ is not continuous at $a$ and $b$. Also, is my rest of the proof correct?
real-analysis proof-verification
real-analysis proof-verification
asked Nov 24 at 5:55
Ashish K
776513
asked Nov 24 at 5:55
Ashish K
776513
edited Nov 24 at 6:02
asked Nov 24 at 5:55
Ashish K
776513
asked Nov 24 at 5:55
Ashish K
776513
asked Nov 24 at 5:55
Ashish K
776513
776513
1
Your proof is fine.The mean value theorem shows that $f$ is strictly increasing on all of the intervals $(a,x_1]$, $[x_k,x_{k+1}]$ and $[x_n,b)$ separately. It is straightforward to show that if $f$ is strictly increasing on $[p_1,p_2]$ and $[p_2,p_3]$ separately then $f$ is strictly increasing on $[p_1,p_3]$.
– copper.hat
Nov 24 at 6:24
@copper.hat it seems I've done a mistake. I've proved that $f(x)<f(x_i)$ but concluded the wrong thing. I need to fix it.
– Ashish K
Nov 24 at 7:04
I don't think you need the first paragraph of your attempt. I'm not sure where you think your mistake is. The proof is not complete, but you have the right idea.
– copper.hat
Nov 24 at 7:32
add a comment |
1
Your proof is fine.The mean value theorem shows that $f$ is strictly increasing on all of the intervals $(a,x_1]$, $[x_k,x_{k+1}]$ and $[x_n,b)$ separately. It is straightforward to show that if $f$ is strictly increasing on $[p_1,p_2]$ and $[p_2,p_3]$ separately then $f$ is strictly increasing on $[p_1,p_3]$.
– copper.hat
Nov 24 at 6:24
@copper.hat it seems I've done a mistake. I've proved that $f(x)<f(x_i)$ but concluded the wrong thing. I need to fix it.
– Ashish K
Nov 24 at 7:04
I don't think you need the first paragraph of your attempt. I'm not sure where you think your mistake is. The proof is not complete, but you have the right idea.
– copper.hat
Nov 24 at 7:32
1
1
Your proof is fine.The mean value theorem shows that $f$ is strictly increasing on all of the intervals $(a,x_1]$, $[x_k,x_{k+1}]$ and $[x_n,b)$ separately. It is straightforward to show that if $f$ is strictly increasing on $[p_1,p_2]$ and $[p_2,p_3]$ separately then $f$ is strictly increasing on $[p_1,p_3]$.
– copper.hat
Nov 24 at 6:24
Your proof is fine.The mean value theorem shows that $f$ is strictly increasing on all of the intervals $(a,x_1]$, $[x_k,x_{k+1}]$ and $[x_n,b)$ separately. It is straightforward to show that if $f$ is strictly increasing on $[p_1,p_2]$ and $[p_2,p_3]$ separately then $f$ is strictly increasing on $[p_1,p_3]$.
– copper.hat
Nov 24 at 6:24
@copper.hat it seems I've done a mistake. I've proved that $f(x)<f(x_i)$ but concluded the wrong thing. I need to fix it.
– Ashish K
Nov 24 at 7:04
@copper.hat it seems I've done a mistake. I've proved that $f(x)<f(x_i)$ but concluded the wrong thing. I need to fix it.
– Ashish K
Nov 24 at 7:04
I don't think you need the first paragraph of your attempt. I'm not sure where you think your mistake is. The proof is not complete, but you have the right idea.
– copper.hat
Nov 24 at 7:32
I don't think you need the first paragraph of your attempt. I'm not sure where you think your mistake is. The proof is not complete, but you have the right idea.
– copper.hat
Nov 24 at 7:32
add a comment |
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1
Your proof is fine.The mean value theorem shows that $f$ is strictly increasing on all of the intervals $(a,x_1]$, $[x_k,x_{k+1}]$ and $[x_n,b)$ separately. It is straightforward to show that if $f$ is strictly increasing on $[p_1,p_2]$ and $[p_2,p_3]$ separately then $f$ is strictly increasing on $[p_1,p_3]$.
– copper.hat
Nov 24 at 6:24
@copper.hat it seems I've done a mistake. I've proved that $f(x)<f(x_i)$ but concluded the wrong thing. I need to fix it.
– Ashish K
Nov 24 at 7:04
I don't think you need the first paragraph of your attempt. I'm not sure where you think your mistake is. The proof is not complete, but you have the right idea.
– copper.hat
Nov 24 at 7:32