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Homework Help: Calculus Help

  1. Oct 16, 2004 #1
    1) Use differentials to estimate f(8.05) given that f(8)=71 and
    (d/dx)f(x)=sqrt of (x+56)

    i used f(c+h) = f(c) + fprime(c) * h
    i got 8.125, but the answer is 71.40

    2) Compute dy/dx at the given values xy - 6y^2 - 2x = -24 , x=1, y=2

    I keep getting 4/25
    but the answer is 0

    3) Give a value c that satisfies the conclusion of the mean value theorem on the interval 2 < x < 5 for the function f(x) = 6x^2 + 8x + 13 (note: all < means greater than or equal to )

    I took the prime of f(x), and f '(c) = 12x + 8
    12x + 8 = 5
    and I got c= (-1/4)
    but the answer is c = (7/4)
     
  2. jcsd
  3. Oct 16, 2004 #2

    Fredrik

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    1. The formula is correct, and when I'm using it, I get 71.4.
    2. The question doesn't make sense.
    3. f'(c) is 50, not 5.
     
    Last edited: Oct 16, 2004
  4. Oct 16, 2004 #3

    cepheid

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    I think 2) should read:

    For the following expression: xy - 6y^2 - 2x = -24

    Compute dy/dx for the following values: x = 1, y = 2

    Differentiate implicitly wrt x:

    [tex] xy - 6y^2 - 2x = -24 [/tex]

    [tex] [x\frac{dy}{dx} + y] - 12y\frac{dy}{dx} - 2 = 0 [/tex]

    [tex] \frac{dy}{dx}(x - 12y) = 2 - y [/tex]

    [tex] \frac{dy}{dx} = \frac{2 - y}{x - 12y} = \frac{2 - 2}{1 - 24} =0 [/tex]
     
  5. Oct 16, 2004 #4

    Fredrik

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    OK, that makes sense. I messed up and thought that (1,2) isn't even on that curve, but I guess 2-24-2 really is -24. :smile:
     
  6. Oct 16, 2004 #5

    HallsofIvy

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    I wish you had shown your work. You seem to be making some serious mistakes. In this case, you are told that f(c)= f(8)= 71 so I don't see how you could haved gotten 71+ something equal to "8.125".
    Yes, f(c+ h)= f(c)+ f'(c)*h (approximately)
    so f(8+ .05)= f(8)+ f'(8)*(0.05)
    = 71+ sqrt(8+56)*(0.05)
    = 71+ sqrt(64)*(0.05)
    = 71+ 8*(0.05)= 71+ 0.4= 71.4
    HOW do you get "4/25"? Using "implicit differentiation",
    y+ xy'- 12yy'- 2= 0. Setting x= 1, y= 2, this is
    2+y'- 24y'- 2 or -23y'= 0. I wonder if you weren't getting the signs mixed up?

    You took the derivative of f(x)! Don't use "slang"!

    The mean value theorem says [itex]\frac{f(b)-f(a)}{b- a}= f '(x)[/itex] for some x between a and b. In this case, a= 2, b= 5 so f(b)= 6(25)+ 8(5)+ 13= 150+ 40+ 13= 203 and f(a)= 6(4)+ 8(2)+ 13= 24+16+ 13= 53. We are looking for c so that [itex]f '(c)= 12c+ 8 = \frac{203-53}{5-2}=\frac{150}{3}= 50. 12c+ 8= 50 so 12c= 42. c= 42/12= 7/2 (?Not 7/4).
    I have no idea where you got "5". If 12c+ 8= 41, then 12c= 33
     
  7. Oct 16, 2004 #6
    Thanks for everyone's help! :smile:
     
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