Find the best upper bound of $varepsilon$ such that $g(x) = 1 + varepsilon f(x) > 0$, $f(x)$ bounded on a...












0












$begingroup$


Consider a smooth function $f : [0, 1] to mathbb{R}$. Moreover, consider:



$$g(x) = 1 + varepsilon f(x),$$



for $x in [0, 1]$ and some $varepsilon > 0$.



Which are the conditions on $varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:



$$0 < varepsilon < M,$$



for some $M > 0$.




My attempt



Let $[0, 1] = I^- cup I^+$, where:





  • $f(x) < 0 ~forall x in I^-$;


  • $f(x) geq 0 ~forall x in I^+$.


For $x in I^+$, then $g(x) > 0$.



For $x in I^-$, since function $f$ is (smooth, then) bounded, then:



$$exists F > 0 : f(x)> -F ~forall x in I^-.$$



Therefore:



$$g(x) = 1 + varepsilon f(x) > 1 - varepsilon F ~forall x in I^-.$$



If $varepsilon < frac{1}{F}$, then:



$$1 - varepsilon F > 0,$$



and hence



$$g(x) > 0 ~forall x in [0, 1].$$




Questions



Is there some easies way to achieve this kind of results?



Is there any better upper bound for $varepsilon$?






algebra-precalculus proof-verification upper-lower-bounds







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Consider a smooth function $f : [0, 1] to mathbb{R}$. Moreover, consider:



    $$g(x) = 1 + varepsilon f(x),$$



    for $x in [0, 1]$ and some $varepsilon > 0$.



    Which are the conditions on $varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:



    $$0 < varepsilon < M,$$



    for some $M > 0$.




    My attempt



    Let $[0, 1] = I^- cup I^+$, where:





    • $f(x) < 0 ~forall x in I^-$;


    • $f(x) geq 0 ~forall x in I^+$.


    For $x in I^+$, then $g(x) > 0$.



    For $x in I^-$, since function $f$ is (smooth, then) bounded, then:



    $$exists F > 0 : f(x)> -F ~forall x in I^-.$$



    Therefore:



    $$g(x) = 1 + varepsilon f(x) > 1 - varepsilon F ~forall x in I^-.$$



    If $varepsilon < frac{1}{F}$, then:



    $$1 - varepsilon F > 0,$$



    and hence



    $$g(x) > 0 ~forall x in [0, 1].$$




    Questions



    Is there some easies way to achieve this kind of results?



    Is there any better upper bound for $varepsilon$?






    algebra-precalculus proof-verification upper-lower-bounds







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Consider a smooth function $f : [0, 1] to mathbb{R}$. Moreover, consider:



      $$g(x) = 1 + varepsilon f(x),$$



      for $x in [0, 1]$ and some $varepsilon > 0$.



      Which are the conditions on $varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:



      $$0 < varepsilon < M,$$



      for some $M > 0$.




      My attempt



      Let $[0, 1] = I^- cup I^+$, where:





      • $f(x) < 0 ~forall x in I^-$;


      • $f(x) geq 0 ~forall x in I^+$.


      For $x in I^+$, then $g(x) > 0$.



      For $x in I^-$, since function $f$ is (smooth, then) bounded, then:



      $$exists F > 0 : f(x)> -F ~forall x in I^-.$$



      Therefore:



      $$g(x) = 1 + varepsilon f(x) > 1 - varepsilon F ~forall x in I^-.$$



      If $varepsilon < frac{1}{F}$, then:



      $$1 - varepsilon F > 0,$$



      and hence



      $$g(x) > 0 ~forall x in [0, 1].$$




      Questions



      Is there some easies way to achieve this kind of results?



      Is there any better upper bound for $varepsilon$?






      algebra-precalculus proof-verification upper-lower-bounds







      share|cite|improve this question









      $endgroup$




      Consider a smooth function $f : [0, 1] to mathbb{R}$. Moreover, consider:



      $$g(x) = 1 + varepsilon f(x),$$



      for $x in [0, 1]$ and some $varepsilon > 0$.



      Which are the conditions on $varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:



      $$0 < varepsilon < M,$$



      for some $M > 0$.




      My attempt



      Let $[0, 1] = I^- cup I^+$, where:





      • $f(x) < 0 ~forall x in I^-$;


      • $f(x) geq 0 ~forall x in I^+$.


      For $x in I^+$, then $g(x) > 0$.



      For $x in I^-$, since function $f$ is (smooth, then) bounded, then:



      $$exists F > 0 : f(x)> -F ~forall x in I^-.$$



      Therefore:



      $$g(x) = 1 + varepsilon f(x) > 1 - varepsilon F ~forall x in I^-.$$



      If $varepsilon < frac{1}{F}$, then:



      $$1 - varepsilon F > 0,$$



      and hence



      $$g(x) > 0 ~forall x in [0, 1].$$




      Questions



      Is there some easies way to achieve this kind of results?



      Is there any better upper bound for $varepsilon$?






      algebra-precalculus proof-verification upper-lower-bounds




      algebra-precalculus proof-verification upper-lower-bounds






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 2 '18 at 16:01









      the_candymanthe_candyman

      8,91632145




      8,91632145






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.

          Then (SPOILER)




          $$
          g(x)geq 0 quad forall xin [0,1] Leftrightarrow
          1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
          $$
          If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
          $varepsilongeq frac{-1}{m}$







          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.



            If $f(x) < 0$, then



            begin{align}
            g(x) > 0
            &iff 1 + epsilon f(x) > 0 \
            &iff epsilon f(x) > -1 \
            &iff epsilon < -dfrac{1}{f(x)}
            end{align}



            So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
              $endgroup$
              – the_candyman
              Dec 2 '18 at 19:30












            • $begingroup$
              I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
              $endgroup$
              – steven gregory
              Dec 2 '18 at 19:40










            • $begingroup$
              In that case, your answer is the same of the one provided by Andrea Cavaliere.
              $endgroup$
              – the_candyman
              Dec 2 '18 at 19:49










            • $begingroup$
              Yes. I just did the logic differenly.
              $endgroup$
              – steven gregory
              Dec 2 '18 at 20:06











            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.

            Then (SPOILER)




            $$
            g(x)geq 0 quad forall xin [0,1] Leftrightarrow
            1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
            $$
            If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
            $varepsilongeq frac{-1}{m}$







            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.

              Then (SPOILER)




              $$
              g(x)geq 0 quad forall xin [0,1] Leftrightarrow
              1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
              $$
              If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
              $varepsilongeq frac{-1}{m}$







              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.

                Then (SPOILER)




                $$
                g(x)geq 0 quad forall xin [0,1] Leftrightarrow
                1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
                $$
                If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
                $varepsilongeq frac{-1}{m}$







                share|cite|improve this answer









                $endgroup$



                It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.

                Then (SPOILER)




                $$
                g(x)geq 0 quad forall xin [0,1] Leftrightarrow
                1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
                $$
                If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
                $varepsilongeq frac{-1}{m}$








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 16:34









                Andrea CavaliereAndrea Cavaliere

                1156




                1156























                    1












                    $begingroup$

                    If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.



                    If $f(x) < 0$, then



                    begin{align}
                    g(x) > 0
                    &iff 1 + epsilon f(x) > 0 \
                    &iff epsilon f(x) > -1 \
                    &iff epsilon < -dfrac{1}{f(x)}
                    end{align}



                    So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
                      $endgroup$
                      – the_candyman
                      Dec 2 '18 at 19:30












                    • $begingroup$
                      I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
                      $endgroup$
                      – steven gregory
                      Dec 2 '18 at 19:40










                    • $begingroup$
                      In that case, your answer is the same of the one provided by Andrea Cavaliere.
                      $endgroup$
                      – the_candyman
                      Dec 2 '18 at 19:49










                    • $begingroup$
                      Yes. I just did the logic differenly.
                      $endgroup$
                      – steven gregory
                      Dec 2 '18 at 20:06
















                    1












                    $begingroup$

                    If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.



                    If $f(x) < 0$, then



                    begin{align}
                    g(x) > 0
                    &iff 1 + epsilon f(x) > 0 \
                    &iff epsilon f(x) > -1 \
                    &iff epsilon < -dfrac{1}{f(x)}
                    end{align}



                    So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
                      $endgroup$
                      – the_candyman
                      Dec 2 '18 at 19:30












                    • $begingroup$
                      I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
                      $endgroup$
                      – steven gregory
                      Dec 2 '18 at 19:40










                    • $begingroup$
                      In that case, your answer is the same of the one provided by Andrea Cavaliere.
                      $endgroup$
                      – the_candyman
                      Dec 2 '18 at 19:49










                    • $begingroup$
                      Yes. I just did the logic differenly.
                      $endgroup$
                      – steven gregory
                      Dec 2 '18 at 20:06














                    1












                    1








                    1





                    $begingroup$

                    If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.



                    If $f(x) < 0$, then



                    begin{align}
                    g(x) > 0
                    &iff 1 + epsilon f(x) > 0 \
                    &iff epsilon f(x) > -1 \
                    &iff epsilon < -dfrac{1}{f(x)}
                    end{align}



                    So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$






                    share|cite|improve this answer











                    $endgroup$



                    If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.



                    If $f(x) < 0$, then



                    begin{align}
                    g(x) > 0
                    &iff 1 + epsilon f(x) > 0 \
                    &iff epsilon f(x) > -1 \
                    &iff epsilon < -dfrac{1}{f(x)}
                    end{align}



                    So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 2 '18 at 19:39

























                    answered Dec 2 '18 at 19:27









                    steven gregorysteven gregory

                    18.2k32258




                    18.2k32258












                    • $begingroup$
                      Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
                      $endgroup$
                      – the_candyman
                      Dec 2 '18 at 19:30












                    • $begingroup$
                      I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
                      $endgroup$
                      – steven gregory
                      Dec 2 '18 at 19:40










                    • $begingroup$
                      In that case, your answer is the same of the one provided by Andrea Cavaliere.
                      $endgroup$
                      – the_candyman
                      Dec 2 '18 at 19:49










                    • $begingroup$
                      Yes. I just did the logic differenly.
                      $endgroup$
                      – steven gregory
                      Dec 2 '18 at 20:06


















                    • $begingroup$
                      Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
                      $endgroup$
                      – the_candyman
                      Dec 2 '18 at 19:30












                    • $begingroup$
                      I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
                      $endgroup$
                      – steven gregory
                      Dec 2 '18 at 19:40










                    • $begingroup$
                      In that case, your answer is the same of the one provided by Andrea Cavaliere.
                      $endgroup$
                      – the_candyman
                      Dec 2 '18 at 19:49










                    • $begingroup$
                      Yes. I just did the logic differenly.
                      $endgroup$
                      – steven gregory
                      Dec 2 '18 at 20:06
















                    $begingroup$
                    Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
                    $endgroup$
                    – the_candyman
                    Dec 2 '18 at 19:30






                    $begingroup$
                    Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
                    $endgroup$
                    – the_candyman
                    Dec 2 '18 at 19:30














                    $begingroup$
                    I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
                    $endgroup$
                    – steven gregory
                    Dec 2 '18 at 19:40




                    $begingroup$
                    I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
                    $endgroup$
                    – steven gregory
                    Dec 2 '18 at 19:40












                    $begingroup$
                    In that case, your answer is the same of the one provided by Andrea Cavaliere.
                    $endgroup$
                    – the_candyman
                    Dec 2 '18 at 19:49




                    $begingroup$
                    In that case, your answer is the same of the one provided by Andrea Cavaliere.
                    $endgroup$
                    – the_candyman
                    Dec 2 '18 at 19:49












                    $begingroup$
                    Yes. I just did the logic differenly.
                    $endgroup$
                    – steven gregory
                    Dec 2 '18 at 20:06




                    $begingroup$
                    Yes. I just did the logic differenly.
                    $endgroup$
                    – steven gregory
                    Dec 2 '18 at 20:06


















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