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$f(0)=0, f'(X)$ is str. increasing: prove that $f(X)/X$ is also increasing function all $X in(0, infty)$ [duplicate]
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Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0
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$f(0)=0, f'(X)$ is a strictly increasing function: prove that $f(X)/X$ is also strictly increasing for all $X in(0, infty)$
calculus
asked Jul 30 at 18:48
prakhar Gupta
11
marked as duplicate by amWhy, dxiv, Henrik, Strants, Rhys Steele Jul 30 at 22:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0
2 answers
$f(0)=0, f'(X)$ is a strictly increasing function: prove that $f(X)/X$ is also strictly increasing for all $X in(0, infty)$
calculus
asked Jul 30 at 18:48
prakhar Gupta
11
marked as duplicate by amWhy, dxiv, Henrik, Strants, Rhys Steele Jul 30 at 22:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Could you please define MSIF, and show what work you've done on the problem?
– Adrian Keister
Jul 30 at 18:50
MSIF is monotonically strictly increasing function .
– prakhar Gupta
Jul 30 at 18:55
I assume f(X)/X as H(X) and differentiate it H'(X) but could able to proof that H'(X) is greater than 0
– prakhar Gupta
Jul 30 at 18:57
What arithmetic error?
– prakhar Gupta
Jul 30 at 19:33
3
Duplicate of Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0. Also of Show that if $f'$ is strictly increasing, then $fracf(x)x$ is increasing over $(0,infty)$.
– dxiv
Jul 30 at 19:37
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up vote
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favorite
This question already has an answer here:
Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0
2 answers
$f(0)=0, f'(X)$ is a strictly increasing function: prove that $f(X)/X$ is also strictly increasing for all $X in(0, infty)$
calculus
asked Jul 30 at 18:48
prakhar Gupta
11
This question already has an answer here:
Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0
2 answers
$f(0)=0, f'(X)$ is a strictly increasing function: prove that $f(X)/X$ is also strictly increasing for all $X in(0, infty)$
This question already has an answer here:
Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0
2 answers
calculus
asked Jul 30 at 18:48
prakhar Gupta
11
edited Jul 30 at 19:58
asked Jul 30 at 18:48
prakhar Gupta
11
asked Jul 30 at 18:48
prakhar Gupta
11
asked Jul 30 at 18:48
prakhar Gupta
11
11
marked as duplicate by amWhy, dxiv, Henrik, Strants, Rhys Steele Jul 30 at 22:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by amWhy, dxiv, Henrik, Strants, Rhys Steele Jul 30 at 22:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Could you please define MSIF, and show what work you've done on the problem?
– Adrian Keister
Jul 30 at 18:50
MSIF is monotonically strictly increasing function .
– prakhar Gupta
Jul 30 at 18:55
I assume f(X)/X as H(X) and differentiate it H'(X) but could able to proof that H'(X) is greater than 0
– prakhar Gupta
Jul 30 at 18:57
What arithmetic error?
– prakhar Gupta
Jul 30 at 19:33
3
Duplicate of Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0. Also of Show that if $f'$ is strictly increasing, then $fracf(x)x$ is increasing over $(0,infty)$.
– dxiv
Jul 30 at 19:37
add a comment |Â
Could you please define MSIF, and show what work you've done on the problem?
– Adrian Keister
Jul 30 at 18:50
MSIF is monotonically strictly increasing function .
– prakhar Gupta
Jul 30 at 18:55
I assume f(X)/X as H(X) and differentiate it H'(X) but could able to proof that H'(X) is greater than 0
– prakhar Gupta
Jul 30 at 18:57
What arithmetic error?
– prakhar Gupta
Jul 30 at 19:33
3
Duplicate of Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0. Also of Show that if $f'$ is strictly increasing, then $fracf(x)x$ is increasing over $(0,infty)$.
– dxiv
Jul 30 at 19:37
Could you please define MSIF, and show what work you've done on the problem?
– Adrian Keister
Jul 30 at 18:50
Could you please define MSIF, and show what work you've done on the problem?
– Adrian Keister
Jul 30 at 18:50
MSIF is monotonically strictly increasing function .
– prakhar Gupta
Jul 30 at 18:55
MSIF is monotonically strictly increasing function .
– prakhar Gupta
Jul 30 at 18:55
I assume f(X)/X as H(X) and differentiate it H'(X) but could able to proof that H'(X) is greater than 0
– prakhar Gupta
Jul 30 at 18:57
I assume f(X)/X as H(X) and differentiate it H'(X) but could able to proof that H'(X) is greater than 0
– prakhar Gupta
Jul 30 at 18:57
What arithmetic error?
– prakhar Gupta
Jul 30 at 19:33
What arithmetic error?
– prakhar Gupta
Jul 30 at 19:33
3
3
Duplicate of Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0. Also of Show that if $f'$ is strictly increasing, then $fracf(x)x$ is increasing over $(0,infty)$.
– dxiv
Jul 30 at 19:37
Duplicate of Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0. Also of Show that if $f'$ is strictly increasing, then $fracf(x)x$ is increasing over $(0,infty)$.
– dxiv
Jul 30 at 19:37
add a comment |Â
3 Answers
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For any $X>0$, $f$ is continuous in $[0,X]$ and differentiable in $(0,X)$. By mean value theorem, there exists $cin (0,X)$ such that $f'(c)= frac f(X)-f(0) X-0=frac f(X) X$
Since $f'$ is strictly increasing and $X>c$, we have $f'(X) > f'(c)$, i.e. $ X f'(X) - f(X)>0$
Consider the function $g(X)=frac f(X) X$
$$g'(X)= frac 1 X^2 (X f'(X) - f(X))>0$$
So derivative of $g$ is strictly positive. So, $g$ is strictly increasing function.
answered Jul 30 at 19:39
Vignesh
34314
Ok I got it . This is absolutely correct
– prakhar Gupta
Jul 30 at 19:48
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Hint: Apply the mean value theorem to $f$ on the interval $[0, x]$, for $x > 0$.
answered Jul 30 at 19:35
youngsmasher
34037
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Alright, so we've got 2 conditions:
- $f(0) = 0$
- $f'(x)$ is strictly increasing
Define $h(x) = fracf(x)x$. Then as you suggested, taking the derivative yields:
$$h'(x) = fracx f'(x) - f(x)x^2$$
We want to show that $h(x) > 0$ for all $x in (0, infty)$. To see this, note that by the fundamental theorem of calculus and condition $1$ we have:
$$f(x) = int_0^x f'(t)dt$$
Then by the mean value theorem, we have that there is some $c in (0,x)$ such that:
$$ int_0^x f'(t)dt = xf'(c)$$
By condition $2$, since $c < x$ we have $f'(c) < f'(x)$. Combining this with the above yeilds:
$$f(x) < xf'(x)$$
And now we can complete our proof! The above inequality gives us that $xf'(x) - f(x) > 0$, and for any positive $x$ we know that $x^2 > 0$. Thus, both the numerator and denominator of $h'(x)$ are positive, and therefore $h'(x) > 0$ for all $x in (0, infty)$.
Therefore $fracf(x)x$ is strictly increasing.
answered Jul 30 at 19:39
Joe
6119
It is also correct .now I understand it perfectly
– prakhar Gupta
Jul 30 at 19:51
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
For any $X>0$, $f$ is continuous in $[0,X]$ and differentiable in $(0,X)$. By mean value theorem, there exists $cin (0,X)$ such that $f'(c)= frac f(X)-f(0) X-0=frac f(X) X$
Since $f'$ is strictly increasing and $X>c$, we have $f'(X) > f'(c)$, i.e. $ X f'(X) - f(X)>0$
Consider the function $g(X)=frac f(X) X$
$$g'(X)= frac 1 X^2 (X f'(X) - f(X))>0$$
So derivative of $g$ is strictly positive. So, $g$ is strictly increasing function.
answered Jul 30 at 19:39
Vignesh
34314
Ok I got it . This is absolutely correct
– prakhar Gupta
Jul 30 at 19:48
add a comment |Â
up vote
1
down vote
For any $X>0$, $f$ is continuous in $[0,X]$ and differentiable in $(0,X)$. By mean value theorem, there exists $cin (0,X)$ such that $f'(c)= frac f(X)-f(0) X-0=frac f(X) X$
Since $f'$ is strictly increasing and $X>c$, we have $f'(X) > f'(c)$, i.e. $ X f'(X) - f(X)>0$
Consider the function $g(X)=frac f(X) X$
$$g'(X)= frac 1 X^2 (X f'(X) - f(X))>0$$
So derivative of $g$ is strictly positive. So, $g$ is strictly increasing function.
answered Jul 30 at 19:39
Vignesh
34314
Ok I got it . This is absolutely correct
– prakhar Gupta
Jul 30 at 19:48
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For any $X>0$, $f$ is continuous in $[0,X]$ and differentiable in $(0,X)$. By mean value theorem, there exists $cin (0,X)$ such that $f'(c)= frac f(X)-f(0) X-0=frac f(X) X$
Since $f'$ is strictly increasing and $X>c$, we have $f'(X) > f'(c)$, i.e. $ X f'(X) - f(X)>0$
Consider the function $g(X)=frac f(X) X$
$$g'(X)= frac 1 X^2 (X f'(X) - f(X))>0$$
So derivative of $g$ is strictly positive. So, $g$ is strictly increasing function.
answered Jul 30 at 19:39
Vignesh
34314
For any $X>0$, $f$ is continuous in $[0,X]$ and differentiable in $(0,X)$. By mean value theorem, there exists $cin (0,X)$ such that $f'(c)= frac f(X)-f(0) X-0=frac f(X) X$
Since $f'$ is strictly increasing and $X>c$, we have $f'(X) > f'(c)$, i.e. $ X f'(X) - f(X)>0$
Consider the function $g(X)=frac f(X) X$
$$g'(X)= frac 1 X^2 (X f'(X) - f(X))>0$$
So derivative of $g$ is strictly positive. So, $g$ is strictly increasing function.
answered Jul 30 at 19:39
Vignesh
34314
answered Jul 30 at 19:39
Vignesh
34314
answered Jul 30 at 19:39
Vignesh
34314
answered Jul 30 at 19:39
Vignesh
34314
34314
Ok I got it . This is absolutely correct
– prakhar Gupta
Jul 30 at 19:48
add a comment |Â
Ok I got it . This is absolutely correct
– prakhar Gupta
Jul 30 at 19:48
Ok I got it . This is absolutely correct
– prakhar Gupta
Jul 30 at 19:48
Ok I got it . This is absolutely correct
– prakhar Gupta
Jul 30 at 19:48
add a comment |Â
up vote
0
down vote
Hint: Apply the mean value theorem to $f$ on the interval $[0, x]$, for $x > 0$.
answered Jul 30 at 19:35
youngsmasher
34037
add a comment |Â
up vote
0
down vote
Hint: Apply the mean value theorem to $f$ on the interval $[0, x]$, for $x > 0$.
answered Jul 30 at 19:35
youngsmasher
34037
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: Apply the mean value theorem to $f$ on the interval $[0, x]$, for $x > 0$.
answered Jul 30 at 19:35
youngsmasher
34037
Hint: Apply the mean value theorem to $f$ on the interval $[0, x]$, for $x > 0$.
answered Jul 30 at 19:35
youngsmasher
34037
answered Jul 30 at 19:35
youngsmasher
34037
answered Jul 30 at 19:35
youngsmasher
34037
answered Jul 30 at 19:35
youngsmasher
34037
34037
add a comment |Â
add a comment |Â
up vote
0
down vote
Alright, so we've got 2 conditions:
- $f(0) = 0$
- $f'(x)$ is strictly increasing
Define $h(x) = fracf(x)x$. Then as you suggested, taking the derivative yields:
$$h'(x) = fracx f'(x) - f(x)x^2$$
We want to show that $h(x) > 0$ for all $x in (0, infty)$. To see this, note that by the fundamental theorem of calculus and condition $1$ we have:
$$f(x) = int_0^x f'(t)dt$$
Then by the mean value theorem, we have that there is some $c in (0,x)$ such that:
$$ int_0^x f'(t)dt = xf'(c)$$
By condition $2$, since $c < x$ we have $f'(c) < f'(x)$. Combining this with the above yeilds:
$$f(x) < xf'(x)$$
And now we can complete our proof! The above inequality gives us that $xf'(x) - f(x) > 0$, and for any positive $x$ we know that $x^2 > 0$. Thus, both the numerator and denominator of $h'(x)$ are positive, and therefore $h'(x) > 0$ for all $x in (0, infty)$.
Therefore $fracf(x)x$ is strictly increasing.
answered Jul 30 at 19:39
Joe
6119
It is also correct .now I understand it perfectly
– prakhar Gupta
Jul 30 at 19:51
add a comment |Â
up vote
0
down vote
Alright, so we've got 2 conditions:
- $f(0) = 0$
- $f'(x)$ is strictly increasing
Define $h(x) = fracf(x)x$. Then as you suggested, taking the derivative yields:
$$h'(x) = fracx f'(x) - f(x)x^2$$
We want to show that $h(x) > 0$ for all $x in (0, infty)$. To see this, note that by the fundamental theorem of calculus and condition $1$ we have:
$$f(x) = int_0^x f'(t)dt$$
Then by the mean value theorem, we have that there is some $c in (0,x)$ such that:
$$ int_0^x f'(t)dt = xf'(c)$$
By condition $2$, since $c < x$ we have $f'(c) < f'(x)$. Combining this with the above yeilds:
$$f(x) < xf'(x)$$
And now we can complete our proof! The above inequality gives us that $xf'(x) - f(x) > 0$, and for any positive $x$ we know that $x^2 > 0$. Thus, both the numerator and denominator of $h'(x)$ are positive, and therefore $h'(x) > 0$ for all $x in (0, infty)$.
Therefore $fracf(x)x$ is strictly increasing.
answered Jul 30 at 19:39
Joe
6119
It is also correct .now I understand it perfectly
– prakhar Gupta
Jul 30 at 19:51
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Alright, so we've got 2 conditions:
- $f(0) = 0$
- $f'(x)$ is strictly increasing
Define $h(x) = fracf(x)x$. Then as you suggested, taking the derivative yields:
$$h'(x) = fracx f'(x) - f(x)x^2$$
We want to show that $h(x) > 0$ for all $x in (0, infty)$. To see this, note that by the fundamental theorem of calculus and condition $1$ we have:
$$f(x) = int_0^x f'(t)dt$$
Then by the mean value theorem, we have that there is some $c in (0,x)$ such that:
$$ int_0^x f'(t)dt = xf'(c)$$
By condition $2$, since $c < x$ we have $f'(c) < f'(x)$. Combining this with the above yeilds:
$$f(x) < xf'(x)$$
And now we can complete our proof! The above inequality gives us that $xf'(x) - f(x) > 0$, and for any positive $x$ we know that $x^2 > 0$. Thus, both the numerator and denominator of $h'(x)$ are positive, and therefore $h'(x) > 0$ for all $x in (0, infty)$.
Therefore $fracf(x)x$ is strictly increasing.
answered Jul 30 at 19:39
Joe
6119
Alright, so we've got 2 conditions:
- $f(0) = 0$
- $f'(x)$ is strictly increasing
Define $h(x) = fracf(x)x$. Then as you suggested, taking the derivative yields:
$$h'(x) = fracx f'(x) - f(x)x^2$$
We want to show that $h(x) > 0$ for all $x in (0, infty)$. To see this, note that by the fundamental theorem of calculus and condition $1$ we have:
$$f(x) = int_0^x f'(t)dt$$
Then by the mean value theorem, we have that there is some $c in (0,x)$ such that:
$$ int_0^x f'(t)dt = xf'(c)$$
By condition $2$, since $c < x$ we have $f'(c) < f'(x)$. Combining this with the above yeilds:
$$f(x) < xf'(x)$$
And now we can complete our proof! The above inequality gives us that $xf'(x) - f(x) > 0$, and for any positive $x$ we know that $x^2 > 0$. Thus, both the numerator and denominator of $h'(x)$ are positive, and therefore $h'(x) > 0$ for all $x in (0, infty)$.
Therefore $fracf(x)x$ is strictly increasing.
answered Jul 30 at 19:39
Joe
6119
answered Jul 30 at 19:39
Joe
6119
answered Jul 30 at 19:39
Joe
6119
answered Jul 30 at 19:39
Joe
6119
6119
It is also correct .now I understand it perfectly
– prakhar Gupta
Jul 30 at 19:51
add a comment |Â
It is also correct .now I understand it perfectly
– prakhar Gupta
Jul 30 at 19:51
It is also correct .now I understand it perfectly
– prakhar Gupta
Jul 30 at 19:51
It is also correct .now I understand it perfectly
– prakhar Gupta
Jul 30 at 19:51
add a comment |Â
Could you please define MSIF, and show what work you've done on the problem?
– Adrian Keister
Jul 30 at 18:50
MSIF is monotonically strictly increasing function .
– prakhar Gupta
Jul 30 at 18:55
I assume f(X)/X as H(X) and differentiate it H'(X) but could able to proof that H'(X) is greater than 0
– prakhar Gupta
Jul 30 at 18:57
What arithmetic error?
– prakhar Gupta
Jul 30 at 19:33
3
Duplicate of Show that $x mapsto fracf(x)x$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0. Also of Show that if $f'$ is strictly increasing, then $fracf(x)x$ is increasing over $(0,infty)$.
– dxiv
Jul 30 at 19:37