Solving the minimum value for resistance












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How to solve the problem by using calculus- local maxima and minima ?(By first derivative test)
I solved it but I can't find the local maximum and minimum points.



my progress






algebra-precalculus







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  • Maybe you should post your progress so far, so we can give you better feedback.
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    2 days ago










  • I posted up there.
    – user218102
    2 days ago










  • Your link is asking for a Google sign-in.
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    2 days ago












  • What about now?
    – user218102
    2 days ago










  • I changed it .Thanks for your response .
    – user218102
    2 days ago
















0














How to solve the problem by using calculus- local maxima and minima ?(By first derivative test)
I solved it but I can't find the local maximum and minimum points.



my progress






algebra-precalculus







share|cite|improve this question









New contributor




user218102 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Maybe you should post your progress so far, so we can give you better feedback.
    – 0x539
    2 days ago










  • I posted up there.
    – user218102
    2 days ago










  • Your link is asking for a Google sign-in.
    – 0x539
    2 days ago












  • What about now?
    – user218102
    2 days ago










  • I changed it .Thanks for your response .
    – user218102
    2 days ago














0












0








0







How to solve the problem by using calculus- local maxima and minima ?(By first derivative test)
I solved it but I can't find the local maximum and minimum points.



my progress






algebra-precalculus







share|cite|improve this question









New contributor




user218102 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











How to solve the problem by using calculus- local maxima and minima ?(By first derivative test)
I solved it but I can't find the local maximum and minimum points.



my progress






algebra-precalculus




algebra-precalculus






share|cite|improve this question









New contributor




user218102 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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share|cite|improve this question




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edited 2 days ago







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asked 2 days ago









user218102user218102

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user218102 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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user218102 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Maybe you should post your progress so far, so we can give you better feedback.
    – 0x539
    2 days ago










  • I posted up there.
    – user218102
    2 days ago










  • Your link is asking for a Google sign-in.
    – 0x539
    2 days ago












  • What about now?
    – user218102
    2 days ago










  • I changed it .Thanks for your response .
    – user218102
    2 days ago


















  • Maybe you should post your progress so far, so we can give you better feedback.
    – 0x539
    2 days ago










  • I posted up there.
    – user218102
    2 days ago










  • Your link is asking for a Google sign-in.
    – 0x539
    2 days ago












  • What about now?
    – user218102
    2 days ago










  • I changed it .Thanks for your response .
    – user218102
    2 days ago
















Maybe you should post your progress so far, so we can give you better feedback.
– 0x539
2 days ago




Maybe you should post your progress so far, so we can give you better feedback.
– 0x539
2 days ago












I posted up there.
– user218102
2 days ago




I posted up there.
– user218102
2 days ago












Your link is asking for a Google sign-in.
– 0x539
2 days ago






Your link is asking for a Google sign-in.
– 0x539
2 days ago














What about now?
– user218102
2 days ago




What about now?
– user218102
2 days ago












I changed it .Thanks for your response .
– user218102
2 days ago




I changed it .Thanks for your response .
– user218102
2 days ago










2 Answers
2






active

oldest

votes


















1














$F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.






share|cite|improve this answer





























    1














    Method $1$



    As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.



    Method $2$



    We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$






    share|cite|improve this answer





















    • Why we get a local maximum point when f ''(X)<0 ?
      – user218102
      2 days ago










    • In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
      – Mostafa Ayaz
      2 days ago













    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    $F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.






    share|cite|improve this answer


























      1














      $F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.






      share|cite|improve this answer
























        1












        1








        1






        $F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.






        share|cite|improve this answer












        $F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        0x5390x539

        1,047316




        1,047316























            1














            Method $1$



            As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.



            Method $2$



            We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$






            share|cite|improve this answer





















            • Why we get a local maximum point when f ''(X)<0 ?
              – user218102
              2 days ago










            • In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
              – Mostafa Ayaz
              2 days ago


















            1














            Method $1$



            As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.



            Method $2$



            We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$






            share|cite|improve this answer





















            • Why we get a local maximum point when f ''(X)<0 ?
              – user218102
              2 days ago










            • In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
              – Mostafa Ayaz
              2 days ago
















            1












            1








            1






            Method $1$



            As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.



            Method $2$



            We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$






            share|cite|improve this answer












            Method $1$



            As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.



            Method $2$



            We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Mostafa AyazMostafa Ayaz

            14.1k3937




            14.1k3937












            • Why we get a local maximum point when f ''(X)<0 ?
              – user218102
              2 days ago










            • In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
              – Mostafa Ayaz
              2 days ago




















            • Why we get a local maximum point when f ''(X)<0 ?
              – user218102
              2 days ago










            • In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
              – Mostafa Ayaz
              2 days ago


















            Why we get a local maximum point when f ''(X)<0 ?
            – user218102
            2 days ago




            Why we get a local maximum point when f ''(X)<0 ?
            – user218102
            2 days ago












            In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
            – Mostafa Ayaz
            2 days ago






            In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
            – Mostafa Ayaz
            2 days ago












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