Find the $x$ at which the local maxima of a function ocours.












1












$begingroup$



The function $f(x) = intlimits_{-1}^{x}t(e^t-1)(t-1)(t-2)^3(t-3)^5 dt$ has a local maxima at $x=?$




First, I differentiated $f(x)$ and found its roots.



That came out to be $x = 0,1,2,3$. Now, one of those numbers, when plugged into $f(x)$, must give the largest value compared to the others.



In our case, since we are integrating, we would need to find the $x$ that minimizes the negative area of the function that we are integrating.



By further work and analyzing the function, I figured out that the answer must be $0$ or $2.$



But, I don't know what to do next?



Any help would be appreciated.






calculus integration functions derivatives definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you actually plug in the values and compare..
    $endgroup$
    – Mustafa Said
    Dec 30 '18 at 10:04










  • $begingroup$
    where do i plug the values in? Do we have f(x) just in x to "plug and compare"?
    $endgroup$
    – Tony
    Dec 30 '18 at 10:11
















1












$begingroup$



The function $f(x) = intlimits_{-1}^{x}t(e^t-1)(t-1)(t-2)^3(t-3)^5 dt$ has a local maxima at $x=?$




First, I differentiated $f(x)$ and found its roots.



That came out to be $x = 0,1,2,3$. Now, one of those numbers, when plugged into $f(x)$, must give the largest value compared to the others.



In our case, since we are integrating, we would need to find the $x$ that minimizes the negative area of the function that we are integrating.



By further work and analyzing the function, I figured out that the answer must be $0$ or $2.$



But, I don't know what to do next?



Any help would be appreciated.






calculus integration functions derivatives definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you actually plug in the values and compare..
    $endgroup$
    – Mustafa Said
    Dec 30 '18 at 10:04










  • $begingroup$
    where do i plug the values in? Do we have f(x) just in x to "plug and compare"?
    $endgroup$
    – Tony
    Dec 30 '18 at 10:11














1












1








1


1



$begingroup$



The function $f(x) = intlimits_{-1}^{x}t(e^t-1)(t-1)(t-2)^3(t-3)^5 dt$ has a local maxima at $x=?$




First, I differentiated $f(x)$ and found its roots.



That came out to be $x = 0,1,2,3$. Now, one of those numbers, when plugged into $f(x)$, must give the largest value compared to the others.



In our case, since we are integrating, we would need to find the $x$ that minimizes the negative area of the function that we are integrating.



By further work and analyzing the function, I figured out that the answer must be $0$ or $2.$



But, I don't know what to do next?



Any help would be appreciated.






calculus integration functions derivatives definite-integrals







share|cite|improve this question











$endgroup$





The function $f(x) = intlimits_{-1}^{x}t(e^t-1)(t-1)(t-2)^3(t-3)^5 dt$ has a local maxima at $x=?$




First, I differentiated $f(x)$ and found its roots.



That came out to be $x = 0,1,2,3$. Now, one of those numbers, when plugged into $f(x)$, must give the largest value compared to the others.



In our case, since we are integrating, we would need to find the $x$ that minimizes the negative area of the function that we are integrating.



By further work and analyzing the function, I figured out that the answer must be $0$ or $2.$



But, I don't know what to do next?



Any help would be appreciated.






calculus integration functions derivatives definite-integrals




calculus integration functions derivatives definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 11:16









user376343

3,9834829




3,9834829










asked Dec 30 '18 at 9:49









TonyTony

1459




1459












  • $begingroup$
    Did you actually plug in the values and compare..
    $endgroup$
    – Mustafa Said
    Dec 30 '18 at 10:04










  • $begingroup$
    where do i plug the values in? Do we have f(x) just in x to "plug and compare"?
    $endgroup$
    – Tony
    Dec 30 '18 at 10:11


















  • $begingroup$
    Did you actually plug in the values and compare..
    $endgroup$
    – Mustafa Said
    Dec 30 '18 at 10:04










  • $begingroup$
    where do i plug the values in? Do we have f(x) just in x to "plug and compare"?
    $endgroup$
    – Tony
    Dec 30 '18 at 10:11
















$begingroup$
Did you actually plug in the values and compare..
$endgroup$
– Mustafa Said
Dec 30 '18 at 10:04




$begingroup$
Did you actually plug in the values and compare..
$endgroup$
– Mustafa Said
Dec 30 '18 at 10:04












$begingroup$
where do i plug the values in? Do we have f(x) just in x to "plug and compare"?
$endgroup$
– Tony
Dec 30 '18 at 10:11




$begingroup$
where do i plug the values in? Do we have f(x) just in x to "plug and compare"?
$endgroup$
– Tony
Dec 30 '18 at 10:11










2 Answers
2






active

oldest

votes


















2












$begingroup$

HINT:



The sign of $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5$$ is $-,+,-,+$ respectively on the intervals $(-1,1),(1,2),(2,3),(3,infty).$ From this you can finish without further calculation.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    You differentiated and found the roots correctly:
    $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5=0 Rightarrow \
    x_0={0,1,2,3};$$
    Method 1. Check the neighborhood of the critical points:
    $$begin{array}{c|c|c|c}
    x_0&f'(x_0-epsilon)&f'(x_0+epsilon)&text{Type of stationary point}\
    hline
    0&-&-&text{inflection}\
    1&-&+&text{local minimum}\
    2&+&-&text{local maximum}\
    3&-&+&text{local minimum}\
    end{array}$$



    Method 2. Check the second (or higher) order derivative at the critical points:
    $$begin{align}f''(0)&=0 text{(inconclusive)}\
    f''(1)&=32(e-1)>0 text{(local minimum)}\
    f''(2)&=0 text{(inconclusive)}\
    f''(3)&=0 text{(inconclusive)}\
    end{align}$$
    Now you can use higher order derivative test:
    $$begin{align}f'''(0)&<0 text{(strictly decreasing inflection point)}\
    f'''(2)&=0, f^{(4)}(2)<0 text{(local maximum)}\
    f'''(3)&=f^{(4)}(3)=f^{(5)}(3)=0, f^{(6)}(3)>0 text{(local minimum)}.end{align}$$
    Note: You don't have to calculate the higher order derivatives completely, but you should focus on the corresponding factor and the resulting sign.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      +1 for displaying the behavior of $f'$ via nice table.
      $endgroup$
      – Paramanand Singh
      Dec 31 '18 at 18:34










    • $begingroup$
      Thank you and Happy New 2019!
      $endgroup$
      – farruhota
      Dec 31 '18 at 19:02










    • $begingroup$
      So, Is the answer 2?
      $endgroup$
      – Tony
      Dec 31 '18 at 21:34










    • $begingroup$
      Yes, $f(2)approx -6965.2$ is the local maximum. See WolframAlpha answer (click on approximate form).
      $endgroup$
      – farruhota
      Jan 1 at 14:59














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    HINT:



    The sign of $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5$$ is $-,+,-,+$ respectively on the intervals $(-1,1),(1,2),(2,3),(3,infty).$ From this you can finish without further calculation.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      HINT:



      The sign of $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5$$ is $-,+,-,+$ respectively on the intervals $(-1,1),(1,2),(2,3),(3,infty).$ From this you can finish without further calculation.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        HINT:



        The sign of $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5$$ is $-,+,-,+$ respectively on the intervals $(-1,1),(1,2),(2,3),(3,infty).$ From this you can finish without further calculation.






        share|cite|improve this answer









        $endgroup$



        HINT:



        The sign of $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5$$ is $-,+,-,+$ respectively on the intervals $(-1,1),(1,2),(2,3),(3,infty).$ From this you can finish without further calculation.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 11:22









        user376343user376343

        3,9834829




        3,9834829























            2












            $begingroup$

            You differentiated and found the roots correctly:
            $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5=0 Rightarrow \
            x_0={0,1,2,3};$$
            Method 1. Check the neighborhood of the critical points:
            $$begin{array}{c|c|c|c}
            x_0&f'(x_0-epsilon)&f'(x_0+epsilon)&text{Type of stationary point}\
            hline
            0&-&-&text{inflection}\
            1&-&+&text{local minimum}\
            2&+&-&text{local maximum}\
            3&-&+&text{local minimum}\
            end{array}$$



            Method 2. Check the second (or higher) order derivative at the critical points:
            $$begin{align}f''(0)&=0 text{(inconclusive)}\
            f''(1)&=32(e-1)>0 text{(local minimum)}\
            f''(2)&=0 text{(inconclusive)}\
            f''(3)&=0 text{(inconclusive)}\
            end{align}$$
            Now you can use higher order derivative test:
            $$begin{align}f'''(0)&<0 text{(strictly decreasing inflection point)}\
            f'''(2)&=0, f^{(4)}(2)<0 text{(local maximum)}\
            f'''(3)&=f^{(4)}(3)=f^{(5)}(3)=0, f^{(6)}(3)>0 text{(local minimum)}.end{align}$$
            Note: You don't have to calculate the higher order derivatives completely, but you should focus on the corresponding factor and the resulting sign.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 for displaying the behavior of $f'$ via nice table.
              $endgroup$
              – Paramanand Singh
              Dec 31 '18 at 18:34










            • $begingroup$
              Thank you and Happy New 2019!
              $endgroup$
              – farruhota
              Dec 31 '18 at 19:02










            • $begingroup$
              So, Is the answer 2?
              $endgroup$
              – Tony
              Dec 31 '18 at 21:34










            • $begingroup$
              Yes, $f(2)approx -6965.2$ is the local maximum. See WolframAlpha answer (click on approximate form).
              $endgroup$
              – farruhota
              Jan 1 at 14:59


















            2












            $begingroup$

            You differentiated and found the roots correctly:
            $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5=0 Rightarrow \
            x_0={0,1,2,3};$$
            Method 1. Check the neighborhood of the critical points:
            $$begin{array}{c|c|c|c}
            x_0&f'(x_0-epsilon)&f'(x_0+epsilon)&text{Type of stationary point}\
            hline
            0&-&-&text{inflection}\
            1&-&+&text{local minimum}\
            2&+&-&text{local maximum}\
            3&-&+&text{local minimum}\
            end{array}$$



            Method 2. Check the second (or higher) order derivative at the critical points:
            $$begin{align}f''(0)&=0 text{(inconclusive)}\
            f''(1)&=32(e-1)>0 text{(local minimum)}\
            f''(2)&=0 text{(inconclusive)}\
            f''(3)&=0 text{(inconclusive)}\
            end{align}$$
            Now you can use higher order derivative test:
            $$begin{align}f'''(0)&<0 text{(strictly decreasing inflection point)}\
            f'''(2)&=0, f^{(4)}(2)<0 text{(local maximum)}\
            f'''(3)&=f^{(4)}(3)=f^{(5)}(3)=0, f^{(6)}(3)>0 text{(local minimum)}.end{align}$$
            Note: You don't have to calculate the higher order derivatives completely, but you should focus on the corresponding factor and the resulting sign.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 for displaying the behavior of $f'$ via nice table.
              $endgroup$
              – Paramanand Singh
              Dec 31 '18 at 18:34










            • $begingroup$
              Thank you and Happy New 2019!
              $endgroup$
              – farruhota
              Dec 31 '18 at 19:02










            • $begingroup$
              So, Is the answer 2?
              $endgroup$
              – Tony
              Dec 31 '18 at 21:34










            • $begingroup$
              Yes, $f(2)approx -6965.2$ is the local maximum. See WolframAlpha answer (click on approximate form).
              $endgroup$
              – farruhota
              Jan 1 at 14:59
















            2












            2








            2





            $begingroup$

            You differentiated and found the roots correctly:
            $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5=0 Rightarrow \
            x_0={0,1,2,3};$$
            Method 1. Check the neighborhood of the critical points:
            $$begin{array}{c|c|c|c}
            x_0&f'(x_0-epsilon)&f'(x_0+epsilon)&text{Type of stationary point}\
            hline
            0&-&-&text{inflection}\
            1&-&+&text{local minimum}\
            2&+&-&text{local maximum}\
            3&-&+&text{local minimum}\
            end{array}$$



            Method 2. Check the second (or higher) order derivative at the critical points:
            $$begin{align}f''(0)&=0 text{(inconclusive)}\
            f''(1)&=32(e-1)>0 text{(local minimum)}\
            f''(2)&=0 text{(inconclusive)}\
            f''(3)&=0 text{(inconclusive)}\
            end{align}$$
            Now you can use higher order derivative test:
            $$begin{align}f'''(0)&<0 text{(strictly decreasing inflection point)}\
            f'''(2)&=0, f^{(4)}(2)<0 text{(local maximum)}\
            f'''(3)&=f^{(4)}(3)=f^{(5)}(3)=0, f^{(6)}(3)>0 text{(local minimum)}.end{align}$$
            Note: You don't have to calculate the higher order derivatives completely, but you should focus on the corresponding factor and the resulting sign.






            share|cite|improve this answer









            $endgroup$



            You differentiated and found the roots correctly:
            $$f^{'}(x)=x(e^x-1)(x-1)(x-2)^3(x-3)^5=0 Rightarrow \
            x_0={0,1,2,3};$$
            Method 1. Check the neighborhood of the critical points:
            $$begin{array}{c|c|c|c}
            x_0&f'(x_0-epsilon)&f'(x_0+epsilon)&text{Type of stationary point}\
            hline
            0&-&-&text{inflection}\
            1&-&+&text{local minimum}\
            2&+&-&text{local maximum}\
            3&-&+&text{local minimum}\
            end{array}$$



            Method 2. Check the second (or higher) order derivative at the critical points:
            $$begin{align}f''(0)&=0 text{(inconclusive)}\
            f''(1)&=32(e-1)>0 text{(local minimum)}\
            f''(2)&=0 text{(inconclusive)}\
            f''(3)&=0 text{(inconclusive)}\
            end{align}$$
            Now you can use higher order derivative test:
            $$begin{align}f'''(0)&<0 text{(strictly decreasing inflection point)}\
            f'''(2)&=0, f^{(4)}(2)<0 text{(local maximum)}\
            f'''(3)&=f^{(4)}(3)=f^{(5)}(3)=0, f^{(6)}(3)>0 text{(local minimum)}.end{align}$$
            Note: You don't have to calculate the higher order derivatives completely, but you should focus on the corresponding factor and the resulting sign.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 31 '18 at 16:24









            farruhotafarruhota

            22.5k2942




            22.5k2942












            • $begingroup$
              +1 for displaying the behavior of $f'$ via nice table.
              $endgroup$
              – Paramanand Singh
              Dec 31 '18 at 18:34










            • $begingroup$
              Thank you and Happy New 2019!
              $endgroup$
              – farruhota
              Dec 31 '18 at 19:02










            • $begingroup$
              So, Is the answer 2?
              $endgroup$
              – Tony
              Dec 31 '18 at 21:34










            • $begingroup$
              Yes, $f(2)approx -6965.2$ is the local maximum. See WolframAlpha answer (click on approximate form).
              $endgroup$
              – farruhota
              Jan 1 at 14:59




















            • $begingroup$
              +1 for displaying the behavior of $f'$ via nice table.
              $endgroup$
              – Paramanand Singh
              Dec 31 '18 at 18:34










            • $begingroup$
              Thank you and Happy New 2019!
              $endgroup$
              – farruhota
              Dec 31 '18 at 19:02










            • $begingroup$
              So, Is the answer 2?
              $endgroup$
              – Tony
              Dec 31 '18 at 21:34










            • $begingroup$
              Yes, $f(2)approx -6965.2$ is the local maximum. See WolframAlpha answer (click on approximate form).
              $endgroup$
              – farruhota
              Jan 1 at 14:59


















            $begingroup$
            +1 for displaying the behavior of $f'$ via nice table.
            $endgroup$
            – Paramanand Singh
            Dec 31 '18 at 18:34




            $begingroup$
            +1 for displaying the behavior of $f'$ via nice table.
            $endgroup$
            – Paramanand Singh
            Dec 31 '18 at 18:34












            $begingroup$
            Thank you and Happy New 2019!
            $endgroup$
            – farruhota
            Dec 31 '18 at 19:02




            $begingroup$
            Thank you and Happy New 2019!
            $endgroup$
            – farruhota
            Dec 31 '18 at 19:02












            $begingroup$
            So, Is the answer 2?
            $endgroup$
            – Tony
            Dec 31 '18 at 21:34




            $begingroup$
            So, Is the answer 2?
            $endgroup$
            – Tony
            Dec 31 '18 at 21:34












            $begingroup$
            Yes, $f(2)approx -6965.2$ is the local maximum. See WolframAlpha answer (click on approximate form).
            $endgroup$
            – farruhota
            Jan 1 at 14:59






            $begingroup$
            Yes, $f(2)approx -6965.2$ is the local maximum. See WolframAlpha answer (click on approximate form).
            $endgroup$
            – farruhota
            Jan 1 at 14:59




















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