Question for proving something is bounded on an interval












1












$begingroup$



Let $f(x)$ = $x^2e^{-x}$. Show that $f$ is bounded on $(0,infty)$




I know that $f$ is only bounded on the interval $I$ if and only if there exists some $M>0$ such that $|f(x)| < M$ for all $x in I$. But how would I go about translating this into a proof?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you allowed to use derivatives?
    $endgroup$
    – D.B.
    Dec 16 '18 at 6:25










  • $begingroup$
    No, that's kinda why I am struggling with this.
    $endgroup$
    – Pablo Tores
    Dec 16 '18 at 6:27












  • $begingroup$
    No need to use derivatives. Limits and Continuity.
    $endgroup$
    – xbh
    Dec 16 '18 at 6:28










  • $begingroup$
    Take a limit. Done.
    $endgroup$
    – Sean Roberson
    Dec 16 '18 at 7:10
















1












$begingroup$



Let $f(x)$ = $x^2e^{-x}$. Show that $f$ is bounded on $(0,infty)$




I know that $f$ is only bounded on the interval $I$ if and only if there exists some $M>0$ such that $|f(x)| < M$ for all $x in I$. But how would I go about translating this into a proof?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you allowed to use derivatives?
    $endgroup$
    – D.B.
    Dec 16 '18 at 6:25










  • $begingroup$
    No, that's kinda why I am struggling with this.
    $endgroup$
    – Pablo Tores
    Dec 16 '18 at 6:27












  • $begingroup$
    No need to use derivatives. Limits and Continuity.
    $endgroup$
    – xbh
    Dec 16 '18 at 6:28










  • $begingroup$
    Take a limit. Done.
    $endgroup$
    – Sean Roberson
    Dec 16 '18 at 7:10














1












1








1





$begingroup$



Let $f(x)$ = $x^2e^{-x}$. Show that $f$ is bounded on $(0,infty)$




I know that $f$ is only bounded on the interval $I$ if and only if there exists some $M>0$ such that $|f(x)| < M$ for all $x in I$. But how would I go about translating this into a proof?






real-analysis







share|cite|improve this question











$endgroup$





Let $f(x)$ = $x^2e^{-x}$. Show that $f$ is bounded on $(0,infty)$




I know that $f$ is only bounded on the interval $I$ if and only if there exists some $M>0$ such that $|f(x)| < M$ for all $x in I$. But how would I go about translating this into a proof?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 6:43









Eevee Trainer

7,67221338




7,67221338










asked Dec 16 '18 at 6:22









Pablo ToresPablo Tores

387




387












  • $begingroup$
    Are you allowed to use derivatives?
    $endgroup$
    – D.B.
    Dec 16 '18 at 6:25










  • $begingroup$
    No, that's kinda why I am struggling with this.
    $endgroup$
    – Pablo Tores
    Dec 16 '18 at 6:27












  • $begingroup$
    No need to use derivatives. Limits and Continuity.
    $endgroup$
    – xbh
    Dec 16 '18 at 6:28










  • $begingroup$
    Take a limit. Done.
    $endgroup$
    – Sean Roberson
    Dec 16 '18 at 7:10


















  • $begingroup$
    Are you allowed to use derivatives?
    $endgroup$
    – D.B.
    Dec 16 '18 at 6:25










  • $begingroup$
    No, that's kinda why I am struggling with this.
    $endgroup$
    – Pablo Tores
    Dec 16 '18 at 6:27












  • $begingroup$
    No need to use derivatives. Limits and Continuity.
    $endgroup$
    – xbh
    Dec 16 '18 at 6:28










  • $begingroup$
    Take a limit. Done.
    $endgroup$
    – Sean Roberson
    Dec 16 '18 at 7:10
















$begingroup$
Are you allowed to use derivatives?
$endgroup$
– D.B.
Dec 16 '18 at 6:25




$begingroup$
Are you allowed to use derivatives?
$endgroup$
– D.B.
Dec 16 '18 at 6:25












$begingroup$
No, that's kinda why I am struggling with this.
$endgroup$
– Pablo Tores
Dec 16 '18 at 6:27






$begingroup$
No, that's kinda why I am struggling with this.
$endgroup$
– Pablo Tores
Dec 16 '18 at 6:27














$begingroup$
No need to use derivatives. Limits and Continuity.
$endgroup$
– xbh
Dec 16 '18 at 6:28




$begingroup$
No need to use derivatives. Limits and Continuity.
$endgroup$
– xbh
Dec 16 '18 at 6:28












$begingroup$
Take a limit. Done.
$endgroup$
– Sean Roberson
Dec 16 '18 at 7:10




$begingroup$
Take a limit. Done.
$endgroup$
– Sean Roberson
Dec 16 '18 at 7:10










3 Answers
3






active

oldest

votes


















1












$begingroup$

You can use the Squeeze Theorem. One possibility is:
$$lim_{x rightarrow infty} e^{-x} leq lim_{x rightarrow infty} x^2e^{-x} leq lim_{x rightarrow infty} e^{x/2}e^{-x}.$$
Note that you can easily show that the upper and lower bound limits are both equal to zero.
You can argue that because your function is continuous, it is bounded on any closed interval $[0,N]$,$N in mathbb{R}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Without using derivatives.



    We know that the function is bounded below by $0$. We know that $ln(x)<dfrac{x}{2}$ on the interval $(0,1)$ since $ln(x)<0$ on that interval.



    In the following animated GIF the curve is the graph of $y=dfrac{1}{x}$, the dark shaded region has area $ln(x)$ for $xge1$ and the orange rectangle upon which it is superimposed has area $dfrac{x}{2}$.



    This constitutes a "proof without words" that



    $$ ln(x)<frac{x}{2}quad text{for }xge1$$



    Proof without words



    Thus we have that $$x^2<e^xquadtext{for }x>0$$
    Multiplying both sides of the inequality by the positive expression $e^{-x}$ yields.



    $$x^2e^{-x}<1quad text{for }x>0$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      One approach is to note $$int_0^infty x^2e^{-x}dx = Gamma(3) < infty$$ where $$Gamma(z) = int_0^infty t^{z-1}e^{-t}dt$$ is the Gamma function. It's not true in general that integrable implies bounded (for instance, consider $int_0^1 x^{-1/2}dx$), but in your case the function $x^2e^{-x}$ is continuous everywhere, and therefore bounded on all (compact) subsets $[0, R]$ for all $R > 0$, in particular in a neighbourhood of zero. The fact the integral converges gives you boundedness in a neighbourhood of infinity. Continuity gives you boundedness everywhere else.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        You can use the Squeeze Theorem. One possibility is:
        $$lim_{x rightarrow infty} e^{-x} leq lim_{x rightarrow infty} x^2e^{-x} leq lim_{x rightarrow infty} e^{x/2}e^{-x}.$$
        Note that you can easily show that the upper and lower bound limits are both equal to zero.
        You can argue that because your function is continuous, it is bounded on any closed interval $[0,N]$,$N in mathbb{R}$.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          You can use the Squeeze Theorem. One possibility is:
          $$lim_{x rightarrow infty} e^{-x} leq lim_{x rightarrow infty} x^2e^{-x} leq lim_{x rightarrow infty} e^{x/2}e^{-x}.$$
          Note that you can easily show that the upper and lower bound limits are both equal to zero.
          You can argue that because your function is continuous, it is bounded on any closed interval $[0,N]$,$N in mathbb{R}$.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            You can use the Squeeze Theorem. One possibility is:
            $$lim_{x rightarrow infty} e^{-x} leq lim_{x rightarrow infty} x^2e^{-x} leq lim_{x rightarrow infty} e^{x/2}e^{-x}.$$
            Note that you can easily show that the upper and lower bound limits are both equal to zero.
            You can argue that because your function is continuous, it is bounded on any closed interval $[0,N]$,$N in mathbb{R}$.






            share|cite|improve this answer









            $endgroup$



            You can use the Squeeze Theorem. One possibility is:
            $$lim_{x rightarrow infty} e^{-x} leq lim_{x rightarrow infty} x^2e^{-x} leq lim_{x rightarrow infty} e^{x/2}e^{-x}.$$
            Note that you can easily show that the upper and lower bound limits are both equal to zero.
            You can argue that because your function is continuous, it is bounded on any closed interval $[0,N]$,$N in mathbb{R}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 16 '18 at 7:08









            D.B.D.B.

            1,26518




            1,26518























                0












                $begingroup$

                Without using derivatives.



                We know that the function is bounded below by $0$. We know that $ln(x)<dfrac{x}{2}$ on the interval $(0,1)$ since $ln(x)<0$ on that interval.



                In the following animated GIF the curve is the graph of $y=dfrac{1}{x}$, the dark shaded region has area $ln(x)$ for $xge1$ and the orange rectangle upon which it is superimposed has area $dfrac{x}{2}$.



                This constitutes a "proof without words" that



                $$ ln(x)<frac{x}{2}quad text{for }xge1$$



                Proof without words



                Thus we have that $$x^2<e^xquadtext{for }x>0$$
                Multiplying both sides of the inequality by the positive expression $e^{-x}$ yields.



                $$x^2e^{-x}<1quad text{for }x>0$$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Without using derivatives.



                  We know that the function is bounded below by $0$. We know that $ln(x)<dfrac{x}{2}$ on the interval $(0,1)$ since $ln(x)<0$ on that interval.



                  In the following animated GIF the curve is the graph of $y=dfrac{1}{x}$, the dark shaded region has area $ln(x)$ for $xge1$ and the orange rectangle upon which it is superimposed has area $dfrac{x}{2}$.



                  This constitutes a "proof without words" that



                  $$ ln(x)<frac{x}{2}quad text{for }xge1$$



                  Proof without words



                  Thus we have that $$x^2<e^xquadtext{for }x>0$$
                  Multiplying both sides of the inequality by the positive expression $e^{-x}$ yields.



                  $$x^2e^{-x}<1quad text{for }x>0$$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Without using derivatives.



                    We know that the function is bounded below by $0$. We know that $ln(x)<dfrac{x}{2}$ on the interval $(0,1)$ since $ln(x)<0$ on that interval.



                    In the following animated GIF the curve is the graph of $y=dfrac{1}{x}$, the dark shaded region has area $ln(x)$ for $xge1$ and the orange rectangle upon which it is superimposed has area $dfrac{x}{2}$.



                    This constitutes a "proof without words" that



                    $$ ln(x)<frac{x}{2}quad text{for }xge1$$



                    Proof without words



                    Thus we have that $$x^2<e^xquadtext{for }x>0$$
                    Multiplying both sides of the inequality by the positive expression $e^{-x}$ yields.



                    $$x^2e^{-x}<1quad text{for }x>0$$






                    share|cite|improve this answer









                    $endgroup$



                    Without using derivatives.



                    We know that the function is bounded below by $0$. We know that $ln(x)<dfrac{x}{2}$ on the interval $(0,1)$ since $ln(x)<0$ on that interval.



                    In the following animated GIF the curve is the graph of $y=dfrac{1}{x}$, the dark shaded region has area $ln(x)$ for $xge1$ and the orange rectangle upon which it is superimposed has area $dfrac{x}{2}$.



                    This constitutes a "proof without words" that



                    $$ ln(x)<frac{x}{2}quad text{for }xge1$$



                    Proof without words



                    Thus we have that $$x^2<e^xquadtext{for }x>0$$
                    Multiplying both sides of the inequality by the positive expression $e^{-x}$ yields.



                    $$x^2e^{-x}<1quad text{for }x>0$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 16 '18 at 8:45









                    John Wayland BalesJohn Wayland Bales

                    14.6k21238




                    14.6k21238























                        0












                        $begingroup$

                        One approach is to note $$int_0^infty x^2e^{-x}dx = Gamma(3) < infty$$ where $$Gamma(z) = int_0^infty t^{z-1}e^{-t}dt$$ is the Gamma function. It's not true in general that integrable implies bounded (for instance, consider $int_0^1 x^{-1/2}dx$), but in your case the function $x^2e^{-x}$ is continuous everywhere, and therefore bounded on all (compact) subsets $[0, R]$ for all $R > 0$, in particular in a neighbourhood of zero. The fact the integral converges gives you boundedness in a neighbourhood of infinity. Continuity gives you boundedness everywhere else.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          One approach is to note $$int_0^infty x^2e^{-x}dx = Gamma(3) < infty$$ where $$Gamma(z) = int_0^infty t^{z-1}e^{-t}dt$$ is the Gamma function. It's not true in general that integrable implies bounded (for instance, consider $int_0^1 x^{-1/2}dx$), but in your case the function $x^2e^{-x}$ is continuous everywhere, and therefore bounded on all (compact) subsets $[0, R]$ for all $R > 0$, in particular in a neighbourhood of zero. The fact the integral converges gives you boundedness in a neighbourhood of infinity. Continuity gives you boundedness everywhere else.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            One approach is to note $$int_0^infty x^2e^{-x}dx = Gamma(3) < infty$$ where $$Gamma(z) = int_0^infty t^{z-1}e^{-t}dt$$ is the Gamma function. It's not true in general that integrable implies bounded (for instance, consider $int_0^1 x^{-1/2}dx$), but in your case the function $x^2e^{-x}$ is continuous everywhere, and therefore bounded on all (compact) subsets $[0, R]$ for all $R > 0$, in particular in a neighbourhood of zero. The fact the integral converges gives you boundedness in a neighbourhood of infinity. Continuity gives you boundedness everywhere else.






                            share|cite|improve this answer









                            $endgroup$



                            One approach is to note $$int_0^infty x^2e^{-x}dx = Gamma(3) < infty$$ where $$Gamma(z) = int_0^infty t^{z-1}e^{-t}dt$$ is the Gamma function. It's not true in general that integrable implies bounded (for instance, consider $int_0^1 x^{-1/2}dx$), but in your case the function $x^2e^{-x}$ is continuous everywhere, and therefore bounded on all (compact) subsets $[0, R]$ for all $R > 0$, in particular in a neighbourhood of zero. The fact the integral converges gives you boundedness in a neighbourhood of infinity. Continuity gives you boundedness everywhere else.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 16 '18 at 23:04









                            RileyRiley

                            1825




                            1825






























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