Show $f(x) >0$ for $x>x_0$ if its $f' >f$ and $f(x_0)=0$











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Let $f: mathbb{R} rightarrow mathbb{R}$ be a differentiable function. Suppose that $f'(x)>f(x)$ for all $x in mathbb{R}$, and $f(x_0)=0$ for some $x_0 in R$. Prove that $f(x)>0$ for all $x>x_0$. As an application of this, show that $ae^x=a+x+x^2/2$.




Here is my attempt:
We know that $lim_{h to 0}{frac{f(x+h)-f(x)}{h}}-f(x)geq 0$. So let $x=x_o+h$, $lim_{h to 0}{frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)geq 0$. So $lim_{h to 0}{frac{f(x_0+h)}{h}}geq 0$. And I'm stuck.






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  • You can use strict inequality.
    – user25959
    Nov 24 at 16:40















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Let $f: mathbb{R} rightarrow mathbb{R}$ be a differentiable function. Suppose that $f'(x)>f(x)$ for all $x in mathbb{R}$, and $f(x_0)=0$ for some $x_0 in R$. Prove that $f(x)>0$ for all $x>x_0$. As an application of this, show that $ae^x=a+x+x^2/2$.




Here is my attempt:
We know that $lim_{h to 0}{frac{f(x+h)-f(x)}{h}}-f(x)geq 0$. So let $x=x_o+h$, $lim_{h to 0}{frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)geq 0$. So $lim_{h to 0}{frac{f(x_0+h)}{h}}geq 0$. And I'm stuck.






calculus







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  • You can use strict inequality.
    – user25959
    Nov 24 at 16:40













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Let $f: mathbb{R} rightarrow mathbb{R}$ be a differentiable function. Suppose that $f'(x)>f(x)$ for all $x in mathbb{R}$, and $f(x_0)=0$ for some $x_0 in R$. Prove that $f(x)>0$ for all $x>x_0$. As an application of this, show that $ae^x=a+x+x^2/2$.




Here is my attempt:
We know that $lim_{h to 0}{frac{f(x+h)-f(x)}{h}}-f(x)geq 0$. So let $x=x_o+h$, $lim_{h to 0}{frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)geq 0$. So $lim_{h to 0}{frac{f(x_0+h)}{h}}geq 0$. And I'm stuck.






calculus







share|cite|improve this question














Let $f: mathbb{R} rightarrow mathbb{R}$ be a differentiable function. Suppose that $f'(x)>f(x)$ for all $x in mathbb{R}$, and $f(x_0)=0$ for some $x_0 in R$. Prove that $f(x)>0$ for all $x>x_0$. As an application of this, show that $ae^x=a+x+x^2/2$.




Here is my attempt:
We know that $lim_{h to 0}{frac{f(x+h)-f(x)}{h}}-f(x)geq 0$. So let $x=x_o+h$, $lim_{h to 0}{frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)geq 0$. So $lim_{h to 0}{frac{f(x_0+h)}{h}}geq 0$. And I'm stuck.






calculus




calculus






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asked Nov 24 at 16:37









mathnoob

1,777322




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  • You can use strict inequality.
    – user25959
    Nov 24 at 16:40


















  • You can use strict inequality.
    – user25959
    Nov 24 at 16:40
















You can use strict inequality.
– user25959
Nov 24 at 16:40




You can use strict inequality.
– user25959
Nov 24 at 16:40










2 Answers
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Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.






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    For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
    $$
    g(x) - g(x_0) = g'(c) (x-x_0)
    $$
    i.e.
    $$
    e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
    $$
    so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.






    share|cite|improve this answer























    • Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
      – user25959
      Nov 24 at 16:56










    • Thank you, I have corrected the answer.
      – Rigel
      Nov 24 at 17:07











    Your Answer





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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

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    active

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    accepted










    Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.






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      up vote
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      accepted










      Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.






      share|cite|improve this answer























        up vote
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        accepted







        up vote
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        down vote



        accepted






        Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.






        share|cite|improve this answer












        Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.







        share|cite|improve this answer












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        answered Nov 24 at 16:48









        xbh

        5,6551522




        5,6551522






















            up vote
            0
            down vote













            For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
            $$
            g(x) - g(x_0) = g'(c) (x-x_0)
            $$
            i.e.
            $$
            e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
            $$
            so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.






            share|cite|improve this answer























            • Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
              – user25959
              Nov 24 at 16:56










            • Thank you, I have corrected the answer.
              – Rigel
              Nov 24 at 17:07















            up vote
            0
            down vote













            For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
            $$
            g(x) - g(x_0) = g'(c) (x-x_0)
            $$
            i.e.
            $$
            e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
            $$
            so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.






            share|cite|improve this answer























            • Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
              – user25959
              Nov 24 at 16:56










            • Thank you, I have corrected the answer.
              – Rigel
              Nov 24 at 17:07













            up vote
            0
            down vote










            up vote
            0
            down vote









            For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
            $$
            g(x) - g(x_0) = g'(c) (x-x_0)
            $$
            i.e.
            $$
            e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
            $$
            so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.






            share|cite|improve this answer














            For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
            $$
            g(x) - g(x_0) = g'(c) (x-x_0)
            $$
            i.e.
            $$
            e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
            $$
            so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 24 at 17:16

























            answered Nov 24 at 16:45









            Rigel

            10.7k11319




            10.7k11319












            • Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
              – user25959
              Nov 24 at 16:56










            • Thank you, I have corrected the answer.
              – Rigel
              Nov 24 at 17:07


















            • Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
              – user25959
              Nov 24 at 16:56










            • Thank you, I have corrected the answer.
              – Rigel
              Nov 24 at 17:07
















            Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
            – user25959
            Nov 24 at 16:56




            Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
            – user25959
            Nov 24 at 16:56












            Thank you, I have corrected the answer.
            – Rigel
            Nov 24 at 17:07




            Thank you, I have corrected the answer.
            – Rigel
            Nov 24 at 17:07


















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