Prove that $f(x)/x$ is increasing












0














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







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    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
















0














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







share|cite|improve this question




















  • 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














0












0








0


2





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 0:27









Bernard

118k639112




118k639112










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














  • 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










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 $0f'(0)-f(0)>0$.






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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 $0f'(0)-f(0)>0$.






    share|cite|improve this answer


























      1














      $(frac {f(x)} x)'=frac {xf'(x)-f(x)} {x^{2}}$ 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 $0f'(0)-f(0)>0$.






      share|cite|improve this answer
























        1












        1








        1






        $(frac {f(x)} x)'=frac {xf'(x)-f(x)} {x^{2}}$ 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 $0f'(0)-f(0)>0$.






        share|cite|improve this answer












        $(frac {f(x)} x)'=frac {xf'(x)-f(x)} {x^{2}}$ 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 $0f'(0)-f(0)>0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 '18 at 0:27









        Kavi Rama Murthy

        50.1k31854




        50.1k31854






























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