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Calculus

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Find any absolute maximum and minimum and local maximum and minimum for the function f(x)=x^3-4x+5 on the interval [0,5). Make sure to prove that these points are max/min values.

THis is what I did.

f'(x)=3x^2-4=0
x= +/- 2/sqrt3
-2/sqrt3 is outside of interval so just use 2/sqrt3 (I think that's considered a critical point)

f(2/sqrt3)= 1.921
f(0) = 5
f(5)=110

I'm not sure what to do with all these numbers. I think since 1.921 is the smallest of the numbers that 2/sqrt 3 is the minimum, but is it the local or absolute.
It looks like 5 is the max, but again local or absolute and can I really include this to be max since it isn't even included in the interval [0,5)

Can someone give me some guidance or show me some steps to solve this.

  • Calculus -

    Functionn has local maximum or minimum where first derivation=0

    If second derivate<0 that is local maximum

    If second derivate>0 that is local minimum

    First derivation=3x^2-4=0

    x=+/- 2/sqroot(3)

    sqroot(3)=1.7320508075688772935274463415059

    x=+/- 2/1.7320508075688772935274463415059=
    1.1547005383792515290182975610039

    Second derivate=6x

    6*1.1547005383792515290182975610039=
    6.9282032302755091741097853660235>0

    Function has local maximum for x=1.1547005383792515290182975610039

    f(x)max=f(1.1547005383792515290182975610039)=
    1.9207985643219959226178731706562

    If you want to see graphs of function in google type:
    functios graphs online

    When you see list of results click on:
    rechneronline.de/function-graphs/

    When page be vopen in blue rectangle type:
    x^2x^3-4x+5

    In Range x-axis from type -4 to 6

    In Range y-axis from type -1 to 9

    Then click option Draw

  • Calculus -

    Sorry:

    Second derivate=6x

    6*1.1547005383792515290182975610039=
    6.9282032302755091741097853660235>0

    Function has local minimum for x=1.1547005383792515290182975610039

    f(x)min=f(1.1547005383792515290182975610039)=
    1.9207985643219959226178731706562

  • Calculus -

    So the local min and max are both 2/sqrt3. what about the absolite values or do the not exists?

    Thanks.

  • Calculus -

    Bosnian,

    Thanks for your help. I just want to make sure I understand. The local min and max are the same f(2/sqrt3) = 1.92

    Are there any absolute maximum or minimum?

    Thanks again.

  • Calculus -

    If you try to see graph of that function you can see that function:

    x^3-4x+5

    havent absolute maximum or absolute minimum.

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