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anonymous

  • one year ago

If f(x)= x^3 + x^2 + x +1, find a number c that satisfies the conclusion of the Mean Value Theorem on the interval [0,4]. (This seems very straight forward, but I keep getting the wrong answer).

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  1. SolomonZelman
    • one year ago
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    I am going to assume you know the Mean Value Theorem. Okay?

  2. anonymous
    • one year ago
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    Correct assumption.

  3. SolomonZelman
    • one year ago
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    You need to: 1) Find f(0) 2) Find f(4) 3) Find the average rate of change betwen (0,f(0)) and (4,f(4)) then we will find point(s) that have the same slope as the one you find in step 3./

  4. SolomonZelman
    • one year ago
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    ( first, do the 1st 3 steps )

  5. anonymous
    • one year ago
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    I did all that. Somehow I'm messing up.

  6. SolomonZelman
    • one year ago
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    ok, what was your slope of the secant from x=0 to x=4?

  7. anonymous
    • one year ago
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    21

  8. SolomonZelman
    • one year ago
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    Ok, f(0)=1 f(4)=85 slope of the secant = (85-1)/(4-0) = 84/4 = 21 Correct!

  9. SolomonZelman
    • one year ago
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    Now, find f`(x)

  10. SolomonZelman
    • one year ago
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    And then set the f`(x)=c, in this case 21, and we will see at which points does the function have an instanteneous slope of 21.

  11. anonymous
    • one year ago
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    I've done that twice. I must be making an error.

  12. SolomonZelman
    • one year ago
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    you have a problem taking the derivative of ` x³+x²+x+1 ` , is that correct ?

  13. anonymous
    • one year ago
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    No. I get 3x^2 + 2x +1

  14. SolomonZelman
    • one year ago
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    yes, that is correct. f`(x)=3x²+2x+1

  15. SolomonZelman
    • one year ago
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    REMEMBER! The derivative of the function, is its (the function's) slope. Now, the slope of the secant in our case is 21. And this is why you need to set f`(x)=21, to see at which points will the derivative of the function (or the function's slope) be =21.

  16. anonymous
    • one year ago
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    Yes, I understand that. That's what I did. I think I must be making some stupid mistake, because I'm getting a different answer than the book.

  17. SolomonZelman
    • one year ago
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    what is the answer in the book?

  18. SolomonZelman
    • one year ago
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    and what answer have you previously been getting?

  19. anonymous
    • one year ago
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    I got x = -3 and x = 7/3. The book says it's (sqrt(61 - 1)/3.

  20. anonymous
    • one year ago
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    oops, -3 can't be right.

  21. SolomonZelman
    • one year ago
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    yes, the book is correct.... :) (well, duh, but I actually did it...)

  22. SolomonZelman
    • one year ago
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    when you set the equation 3x²+2x+1=21, you will get two solutions, and you will exclude one of them because this another solution is NOT on the interval of [0,4]. Can you solve 3x²+2x+1=21 ? (Shouldn't be a problem for you))

  23. SolomonZelman
    • one year ago
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    you can use the quadratic formula (i would advise)

  24. anonymous
    • one year ago
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    You would think so, but apparently not.

  25. SolomonZelman
    • one year ago
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    Ok, should I do this algebraic task for you?

  26. SolomonZelman
    • one year ago
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    well, we are getting the calculus algorithm, so that wouldn't do much harm....

  27. anonymous
    • one year ago
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    Well, that seems to be where my brain is malfunctioning.

  28. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle 3x^2+2x+1=21 }\) \(\large\color{black}{ \displaystyle 3x^2+2x-20=0 }\) \(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{2^2-4(3)(-20)}}{2\cdot 3} }\) \(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{224}}{2\cdot 3} }\) \(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{61\cdot 4}}{2\cdot 3} }\) \(\large\color{black}{ \displaystyle \frac{-2\pm 2\sqrt{61}}{2\cdot 3} }\)

  29. SolomonZelman
    • one year ago
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    from here you can proceed with no probs....

  30. anonymous
    • one year ago
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    OMG! Dyslexia strikes again. Instead of putting in 2^2, i put in 4 AND THEN squared that. Thanks so much!! This kind of thing makes me crazy.

  31. SolomonZelman
    • one year ago
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    oh, it is just a minor mistake... messes everything up but happens to everybody:)

  32. anonymous
    • one year ago
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    How do you get the math characters in the display like that?

  33. SolomonZelman
    • one year ago
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    I am using "latex"

  34. anonymous
    • one year ago
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    OK, thanks. I'll have to research that some time. Thanks again.

  35. SolomonZelman
    • one year ago
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    you can look up a "legendary latex tutorial" online, and you should find a tutorial made by thomaster

  36. anonymous
    • one year ago
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    Will do. Thanks.

  37. SolomonZelman
    • one year ago
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    If you want, we can do a polynomial approximation of √61, using f(x)=√x, for x=64.

  38. SolomonZelman
    • one year ago
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    but that is not required I guess....

  39. SolomonZelman
    • one year ago
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    so, your only answer (that is in the interval [0,4] ) = \(\large\color{black}{ \displaystyle \frac{1 }{3} (\sqrt{61}-1) }\)

  40. SolomonZelman
    • one year ago
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    any questions ?

  41. SolomonZelman
    • one year ago
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    So (extra part) you have a function: \(\large\color{black}{ \displaystyle f(x)=\sqrt{x} }\) \(\large\color{black}{ \displaystyle L(x)=f(a)+f`(a)(x-a) }\) this is a linear approximation of the function f(x) near the a point x=a. in this case our function is f(x)=√x, and a=64. (our f`(x) is going to be 1/(2√x), and f`(a)=f`(8)=1/(2√64) ...) \(\large\color{black}{ \displaystyle L(x)=\sqrt{64}+\frac{1}{2\sqrt{64}}(x-64) }\) \(\large\color{black}{ \displaystyle L(x)=8+\frac{1}{16}(x-64) }\) this is the tangent line to √x, at x=64 now, you can plug in 61 to get an approzimate value for √61 \(\large\color{black}{ \displaystyle \sqrt{61}\approx 8+\frac{-3}{16} }\) \(\large\color{black}{ \displaystyle \sqrt{61}\approx 7+\frac{13}{16} }\)

  42. SolomonZelman
    • one year ago
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    But, by taylor approximation, you would have a closer approximation. (that is, drawing a tangent the the f(x) polynomial of nth degree thorugh the point x=a) \(\color{black}{ \displaystyle {\rm Approximation~of~f(x)}_{\rm~at~~x~=~a}~~ \\[1.3em] \displaystyle =f(a)+f`(a)(x-a)+\frac{f``(a)}{2!}(x-a)^2+\frac{f```(a)}{3!}(x-a)^2+ \\[1.3em] \displaystyle {\bf ....}~+~+\frac{f^{(n)}(a)}{n!}(x-a)^n }\)

  43. SolomonZelman
    • one year ago
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    in any case... won't overwhelm more than what I did already

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