Mixing
Prove that $f(x)/x$ is increasing
Suppose $f(0)<0$ and that for all $x$, $f’’(x)>0$. Show that $f(x)/x$ is increasing on $(0,infty)$.
I tried to use the Taylor formula but it didn’t work. Thanks in advance for your help!
calculus analysis
edited Nov 20 '18 at 0:27
Bernard
118k639112
asked Nov 20 '18 at 0:21
Jiu
435111
add a comment |
Suppose $f(0)<0$ and that for all $x$, $f’’(x)>0$. Show that $f(x)/x$ is increasing on $(0,infty)$.
I tried to use the Taylor formula but it didn’t work. Thanks in advance for your help!
calculus analysis
edited Nov 20 '18 at 0:27
Bernard
118k639112
asked Nov 20 '18 at 0:21
Jiu
435111
2
The question in the title doesn't match with the one in the body of the question.
– Kavi Rama Murthy
Nov 20 '18 at 0:24
@KaviRamaMurthy thanks for reminding me
– Jiu
Nov 20 '18 at 0:24
@Bernard sorry I made a mistake in my questions
– Jiu
Nov 20 '18 at 0:25
add a comment |
Suppose $f(0)<0$ and that for all $x$, $f’’(x)>0$. Show that $f(x)/x$ is increasing on $(0,infty)$.
I tried to use the Taylor formula but it didn’t work. Thanks in advance for your help!
calculus analysis
edited Nov 20 '18 at 0:27
Bernard
118k639112
asked Nov 20 '18 at 0:21
Jiu
435111
Suppose $f(0)<0$ and that for all $x$, $f’’(x)>0$. Show that $f(x)/x$ is increasing on $(0,infty)$.
I tried to use the Taylor formula but it didn’t work. Thanks in advance for your help!
calculus analysis
calculus analysis
edited Nov 20 '18 at 0:27
Bernard
118k639112
asked Nov 20 '18 at 0:21
Jiu
435111
edited Nov 20 '18 at 0:27
Bernard
118k639112
asked Nov 20 '18 at 0:21
Jiu
435111
edited Nov 20 '18 at 0:27
Bernard
118k639112
edited Nov 20 '18 at 0:27
Bernard
118k639112
edited Nov 20 '18 at 0:27
Bernard
118k639112
118k639112
asked Nov 20 '18 at 0:21
Jiu
435111
asked Nov 20 '18 at 0:21
Jiu
435111
asked Nov 20 '18 at 0:21
Jiu
435111
435111
2
The question in the title doesn't match with the one in the body of the question.
– Kavi Rama Murthy
Nov 20 '18 at 0:24
@KaviRamaMurthy thanks for reminding me
– Jiu
Nov 20 '18 at 0:24
@Bernard sorry I made a mistake in my questions
– Jiu
Nov 20 '18 at 0:25
add a comment |
2
The question in the title doesn't match with the one in the body of the question.
– Kavi Rama Murthy
Nov 20 '18 at 0:24
@KaviRamaMurthy thanks for reminding me
– Jiu
Nov 20 '18 at 0:24
@Bernard sorry I made a mistake in my questions
– Jiu
Nov 20 '18 at 0:25
2
2
The question in the title doesn't match with the one in the body of the question.
– Kavi Rama Murthy
Nov 20 '18 at 0:24
The question in the title doesn't match with the one in the body of the question.
– Kavi Rama Murthy
Nov 20 '18 at 0:24
@KaviRamaMurthy thanks for reminding me
– Jiu
Nov 20 '18 at 0:24
@KaviRamaMurthy thanks for reminding me
– Jiu
Nov 20 '18 at 0:24
@Bernard sorry I made a mistake in my questions
– Jiu
Nov 20 '18 at 0:25
@Bernard sorry I made a mistake in my questions
– Jiu
Nov 20 '18 at 0:25
add a comment |
1 Answer
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$(frac {f(x)} x)'=frac {xf'(x)-f(x)} {x^{2}}$ so it is enough to show that $xf'(x)-f(x) > 0$. Now $(xf'(x)-f(x))'=xf''(x) > 0$ so $xf'(x)-f(x)$ is increasing and it is enough to observe that $0f'(0)-f(0)>0$.
answered Nov 20 '18 at 0:27
Kavi Rama Murthy
50.1k31854
add a comment |
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1 Answer
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$(frac {f(x)} x)'=frac {xf'(x)-f(x)} {x^{2}}$ so it is enough to show that $xf'(x)-f(x) > 0$. Now $(xf'(x)-f(x))'=xf''(x) > 0$ so $xf'(x)-f(x)$ is increasing and it is enough to observe that $0f'(0)-f(0)>0$.
answered Nov 20 '18 at 0:27
Kavi Rama Murthy
50.1k31854
add a comment |
$(frac {f(x)} x)'=frac {xf'(x)-f(x)} {x^{2}}$ so it is enough to show that $xf'(x)-f(x) > 0$. Now $(xf'(x)-f(x))'=xf''(x) > 0$ so $xf'(x)-f(x)$ is increasing and it is enough to observe that $0f'(0)-f(0)>0$.
answered Nov 20 '18 at 0:27
Kavi Rama Murthy
50.1k31854
add a comment |
$(frac {f(x)} x)'=frac {xf'(x)-f(x)} {x^{2}}$ so it is enough to show that $xf'(x)-f(x) > 0$. Now $(xf'(x)-f(x))'=xf''(x) > 0$ so $xf'(x)-f(x)$ is increasing and it is enough to observe that $0f'(0)-f(0)>0$.
answered Nov 20 '18 at 0:27
Kavi Rama Murthy
50.1k31854
$(frac {f(x)} x)'=frac {xf'(x)-f(x)} {x^{2}}$ so it is enough to show that $xf'(x)-f(x) > 0$. Now $(xf'(x)-f(x))'=xf''(x) > 0$ so $xf'(x)-f(x)$ is increasing and it is enough to observe that $0f'(0)-f(0)>0$.
answered Nov 20 '18 at 0:27
Kavi Rama Murthy
50.1k31854
answered Nov 20 '18 at 0:27
Kavi Rama Murthy
50.1k31854
answered Nov 20 '18 at 0:27
Kavi Rama Murthy
50.1k31854
answered Nov 20 '18 at 0:27
Kavi Rama Murthy
50.1k31854
50.1k31854
add a comment |
add a comment |
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2
The question in the title doesn't match with the one in the body of the question.
– Kavi Rama Murthy
Nov 20 '18 at 0:24
@KaviRamaMurthy thanks for reminding me
– Jiu
Nov 20 '18 at 0:24
@Bernard sorry I made a mistake in my questions
– Jiu
Nov 20 '18 at 0:25