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zeesbrat3

  • one year ago

If f(x) = ∣(x2 − 8)∣, how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem?

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  1. zeesbrat3
    • one year ago
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    Please help

  2. amoodarya
    • one year ago
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    if u plot it , it will be simpler to do

  3. amoodarya
    • one year ago
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    \[|x^2-8|=|(x-2\sqrt{2})(x+2\sqrt{2})|\]

  4. amoodarya
    • one year ago
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    |dw:1436664302388:dw|

  5. amoodarya
    • one year ago
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    |dw:1436664349288:dw|

  6. amoodarya
    • one year ago
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    |dw:1436664378196:dw|

  7. zeesbrat3
    • one year ago
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    So, 2 because those are the 0s?

  8. zeesbrat3
    • one year ago
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    @amoodarya

  9. amoodarya
    • one year ago
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    can you tell me what kind of point , theorem want ?

  10. zeesbrat3
    • one year ago
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    You're looking for a point that is defined and continuous on the supplied interval as well as differentiable... right?

  11. amoodarya
    • one year ago
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    yes

  12. misty1212
    • one year ago
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    HI!! (and holy moly!!0

  13. misty1212
    • one year ago
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    you are not looking for a point that is defined and continuous, that doesn't even make any sense !!

  14. misty1212
    • one year ago
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    first you have to check to see if your function is differentiable on the interval you are given

  15. amoodarya
    • one year ago
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    |dw:1436664956891:dw| does it make sense ?

  16. misty1212
    • one year ago
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    since \[|x^2-8|=-x^2+8\] on your interval, it is definitely continuous and differentiable (because it is a polynomial)

  17. misty1212
    • one year ago
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    you are not asked do find any number here, just how many numbers will satisfy the conclusion of the mean value theorem

  18. misty1212
    • one year ago
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    since your function is a quadratic, the derivative is a line when you solve a linear equation there is only one solution

  19. misty1212
    • one year ago
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    in fact you can check that the solution to the number guarantee to exist by the mean value theorem is in the center of the interval that is always the case with a quadratic

  20. zeesbrat3
    • one year ago
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    Thank you for your explanation!

  21. misty1212
    • one year ago
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    you are welcome, i hope it is clear (more or less) and to reiterate (not to be mean) "You're looking for a point that is defined and continuous on the supplied interval as well as differentiable... right? makes zero sense a "point" is a number a number is always defined, it is a number a number is not "continuous" that is a local property of a function, not a number nor is a number "differentiable" a function may or may not be differentiable on an interval, i.e. the derivative may or may not exist for all numbers in the interval

  22. ybarrap
    • one year ago
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    If a function f is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that $$ f'(c) = \frac{f(b) - f(a)}{b-a} $$ |dw:1436665525899:dw| https://en.wikipedia.org/wiki/Mean_value_theorem https://www.wolframalpha.com/input/?i=plot+-x%5E2+%2B+8%2C0%3C%3Dx%3C%3D2.5

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