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zeesbrat3

  • one year ago

Which of the following statements is/are true? a.) If f '(c) = 0, then f has a local maximum or minimum at x = c. b.) If f is continuous on [a, b] and differentiable on (a, b) and f '(x) = 0 on (a, b), then f is constant on [a, b]. c.) The Mean Value Theorem can be applied to f(x) = 1/x2 on the interval [−1, 1].

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  1. zeesbrat3
    • one year ago
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    @ganeshie8 @mathstudent55 @kropot72 @Kash_TheSmartGuy

  2. zeesbrat3
    • one year ago
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    Please help

  3. Michele_Laino
    • one year ago
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    hint: for last statement, we have that x=0 belongs to interval [-1,1], and at x=0, the functio f(x)= 1/x^2 is not continuous

  4. zeesbrat3
    • one year ago
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    Because it wouldnt exist?

  5. Michele_Laino
    • one year ago
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    since we can not divide by zero

  6. zeesbrat3
    • one year ago
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    Sorry, I thought of limits when I said it doesn't exist since like you said, you can't divide by 0

  7. zeesbrat3
    • one year ago
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    I think A i incorrect as well

  8. Michele_Laino
    • one year ago
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    yes! I think so, since we can have this case: |dw:1437332734543:dw|

  9. zeesbrat3
    • one year ago
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    I did it algebraically, but got the same thing. I figured if you had f(x) = x^3 then you wouldn't get a max or min anyways, just a hip

  10. zeesbrat3
    • one year ago
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    So, since A and C are incorrect, the only valid answer would be B, yes?

  11. Michele_Laino
    • one year ago
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    another counterexample is given by this function: f(x)= x^3 we have: f '(x) = 3x^2 and f '(0)==, nevertheless x=0 is not a point of local maximum or minimum |dw:1437333090875:dw|

  12. Michele_Laino
    • one year ago
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    oops.. f '(0)=0

  13. Michele_Laino
    • one year ago
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    yes! option B is the correct one

  14. zeesbrat3
    • one year ago
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    Thank you for clarifying C for me. :) Would you mind helping me with another?

  15. Michele_Laino
    • one year ago
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    ok!

  16. zeesbrat3
    • one year ago
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    Given f '(x) = (x − 4)(4 − 2x), find the x-coordinate for the relative maximum on the graph of f(x).

  17. Michele_Laino
    • one year ago
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    In general when I have to solve a problem like this, I develop the multiplications, namely I rewrite your function as follows: \[f\left( x \right) = - 2{x^2} + 12x - 16\]

  18. Michele_Laino
    • one year ago
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    then I solve this inequality: \[ - 2{x^2} + 12x - 16 \geqslant 0\]

  19. Michele_Laino
    • one year ago
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    which is equivalent to this one: \[{x^2} - 6x + 8 \leqslant 0\]

  20. zeesbrat3
    • one year ago
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    I got \[-2(x-4)(x-2)\ge0\]

  21. Michele_Laino
    • one year ago
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    yes! it is right!

  22. zeesbrat3
    • one year ago
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    Then divide both sides by -2. So 4 is the max?

  23. zeesbrat3
    • one year ago
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    Because it is the highest point

  24. Michele_Laino
    • one year ago
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    we have to proceed with caution

  25. Michele_Laino
    • one year ago
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    the solution of our inequlity is: \[2 \leqslant x \leqslant 4\]

  26. Michele_Laino
    • one year ago
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    inequality*

  27. Michele_Laino
    • one year ago
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    so, we have thi drawing: |dw:1437333653697:dw|

  28. Michele_Laino
    • one year ago
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    where f '(x) is positive then f(x) is increasing where f '(x ) is negative then f(x) is decreasing

  29. zeesbrat3
    • one year ago
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    The number line is your checks to see whether it is min or max..

  30. Michele_Laino
    • one year ago
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    yes!

  31. Michele_Laino
    • one year ago
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    |dw:1437333827000:dw|

  32. perl
    • one year ago
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    Given f '(x) = (x − 4)(4 − 2x), find the x-coordinate for the relative maximum on the graph of f(x). The x coordinate of rel. max/min occurs when f ' (x) = 0 note this is a necessary condition, not a sufficient

  33. Michele_Laino
    • one year ago
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    as we can see from my drawing x=4 is a point of local maximum and x=2 is a point of local minimum

  34. zeesbrat3
    • one year ago
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    Right, now what if you wanted to use that graph to find the max of the original function f(x)?

  35. Michele_Laino
    • one year ago
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    the usage of my graph above is only to know that at x=4 there is a point of local maximum

  36. Michele_Laino
    • one year ago
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    and, of course at x=2 there is a point of local minimum

  37. zeesbrat3
    • one year ago
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    Okay, if we took a graph, like this one: How would you find the max and min of the original function?

  38. Michele_Laino
    • one year ago
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    it is simple, since from your graph we have three stationary points for f(x), furthermore only at x=1 there is a local maximum, since for point at left of x=1 the derivative is positive, so the function is incresing, whereas for point at right of x=1 the derivative is negative, so the function is decreasing. In other words there exists a neighborhood of x=1, such that the function is increasing at the left of x=1 and it is increasing at the right of x=1

  39. Michele_Laino
    • one year ago
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    increasing*

  40. Michele_Laino
    • one year ago
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    |dw:1437334409653:dw|

  41. zeesbrat3
    • one year ago
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    So 1 would be the max of the original function? Is there any way to find the min as well, or is there not enough information?

  42. Michele_Laino
    • one year ago
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    the minimum points are at x=0 and x=2, since for each of those points we have a neighborhood which contains point for which f is decreasing at the left, and points at which f is increasing at the right of the point which the neighborhood is referring to

  43. zeesbrat3
    • one year ago
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    Oh, thank you! I understand now. I wish I was able to give you more than one medal!

  44. Michele_Laino
    • one year ago
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    |dw:1437334708640:dw|

  45. Michele_Laino
    • one year ago
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    thanks! :)

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