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Differentiation -Help, please!

  1. Apr 13, 2008 #1
    Differentiation ---Help, please!

    1. The problem statement, all variables and given/known data
    1) Use Newtons Method to approximate the real zero of the function f(x)=x^3 + 7x + 3 = 0

    2) The curve y = x^3-2x^2+x-3 intersects the curve y = cos(2x). If x = 1.8 is used as the first estimate then, using Newton's Method, what is the x-value or the intersection point, accurate to 4 decimal places?

    3) A curve is given by y = a^3, where a = m^2(2m-1). Approximate the slope of the curve when m = 0.6543

    4) If y = -4 / [tex]\sqrt[3]{x+5}[/tex], then dy/dx

    5) Find an equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1

    2. Relevant equations
    1) Newtons Method:
    x = f(x) / f'(x)
    Or Slope = Rise/Run

    2) Newtons Method
    x = f(x) / f'(x)
    Or Slope = Rise/Run

    3) y = mx+b
    derivatives

    4) Quotient Rule
    (f'g-g'f) / (g^2)

    5) Quotient Rule
    (f'g-g'f) / (g^2)
    y=mx+b

    3. The attempt at a solution
    1) x^3 + 7x = 0
    Let the "guess" be "n"
    Let the next "guess" to be "n2" (2 was suppose to be a subscript but I can't find it in the Latex Reference)
    Slope: derivative of x^3 + 7x = 0
    3x^2 + 7 = 0
    Rise: Plug in the "n" to the equation
    n^3 + 7n = 0
    Run: "n" subtract the "n2"
    n - n2
    Final formula:
    3x^2 + 7 = n^3 + 7n / n - n2
    *The answer should be this: (I was close but my rise is really off)
    The first "n" is suppose to be a lower subscript. I can't find it here in the Latex Reference.
    xn +1 = 2x[tex]^{3}_{n}[/tex] -3 / 3x[tex]^{2}_{n}[/tex] + 7

    2) x^3 - 2x^2 + x - 3 - cos(2x) = 0
    Slope = Rise/Run
    Slope: Derivative of x^3 - 2x^2 + x - 3 - cos(2x)
    3(1.8)^2 - 4(1.8) + 1 + sin2(1.8)(2)
    Rise: Plug in 1.8 to the equation
    (1.8)^3-2(1.8)^2 + 1.8 -3 - cos(2(1.8))
    Run: 1.8 - x
    Final Equation:
    3(1.8)^2 - 4(1.8) + 1 + sin2(1.8)(2) = (1.8)^3-2(1.8)^2 + 1.8 -3 - cos(2(1.8)) / 1.8 - x
    Simplify: 3.645639188 = -2.846026728 / 1.8 - x
    -2.846026728 / 3.645639188 = 1.8 - x
    1.8 + 0.780666045 = x
    x = 2.680666046
    The answer in the book is 2.0923.. So, do you think I should use my "x" to be the second trial and do the whole equation again from the start to get the 2.0923?

    3)
    a = m^2(2m-1)
    (0.6543)^2(2(0.6543)-1))
    a= 0.13211428
    y = a^3
    Therefore...
    y = 0.13211428^3
    y =0.002305946

    I'm stuck. Please help?

    4) ((0)([tex]\sqrt[3]{x+5}[/tex])) - ((-4)(1/3(x+5)^-2/3)) / ([tex]\sqrt[3]{x+5}[/tex])^2)
    Simplify: ((4)(1/3(x+5)^-2/3) / ([tex]\sqrt[3]{x+5}[/tex])
    Then I don't know how to simplify it much longer.. My simplifying skills has gone bad, I guess. Help, please?

    5) Get y-value, plug in the x-value which is 1
    1-1 / 1+1 = 0
    (1,0)
    Do quotient rule on the equation...
    (1)(x+1) - (x-1)(1) / (x+1)^2
    Simplify...
    (x+1) - (x-1) / (x^2 + 2x+2)
    Plug in the x-value which is 1 to get the slope.
    (1+1) - (1-1) / (1^2 +2(1) + 2)
    2 / 1+2+2 = 2 / 5
    Slope = 2/5

    y=mx+b
    0 = 2/5(1) + b
    0 - 2/5 = b
    -2/5 = b
    Final equation...
    y = 2/5x -2/5

    Although it seems like I did everything right, my final equation seems to be incorrect from the answer in the book which is x-2y=1





    THAT'S ALL FOR NOW. I'll continue more questions later on if I find any that bugs my mind. But for now, I need to go to church so in the meanwhile, PLEASE HELP? =/
     
  2. jcsd
  3. Apr 13, 2008 #2
    Anyone there?
     
  4. Apr 14, 2008 #3
    Please?
     
  5. Apr 14, 2008 #4
    At least just help me with question #3.. PLEASE?
     
  6. Apr 14, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That's the definition of "slope", not "Newtons method". Newton's method says [itex]x_{n+1}= x_n- f(x_n)/f '(x_n)[/itex]

    [/itex]2) Newtons Method
    x = f(x) / f'(x)
    Or Slope = Rise/Run
    You understand, don't you, that a "real zero of the function f(x)=x^3 + 7x + 3 = 0" is a number? Applied to this problem, f(x)= x3+ 7x+ 3 so the derivative is 3x2+ 7. Newton's method says that if xn is a close to a solution then [itex]x_{n+1}= x_n- (x^3+ 7x+ 3)/(3x^3+ 7)[/itex] will be closer. If you take x0= 0, then [itex]x_1= 0- (3/7)= -3/7[/itex] or about -0.4286, then [itex]x_2= -0.4181[/itex], etc.

    I think what you are doing is using the idea behind Newton's method rather than Newton's method itself: Yes, you are trying to solve [itex]f(x)= x^3- 2x^2+ x- 3- cos(2x)= 0[/itex]. [itex]f '(x)= 3x^2- 4x+ 1+ 2sin(2x)[/itex]. If you choose 1.8 as your x0, then [itex]x_1= 1.8- ((1.8)^3- 2(1.8)^2+ 1.8- 3- cos(3.6))/(3(1.8)^2- 4(1.8)+ 1+ sin(3.6)= 2.1091[/itex], not the "2.6807" you got. Try it again and then continue until you get two consecutive values that are equal to 4 decimal places.

    You weren't asked to find y, you were asked to find the derivative of a: use the chain rule: dy/dm= (dy/da)(da/dm). To find da/dm, I think I would go ahead and multiply: a= m2(2m-1)= m3- m2

    I think you will find it much easier to write y= -4(x+ 5)-1/3 and use the 'power rule' rather than using the quotient rule.

    No, (x+1)2 is NOT "x2+ 2x+ 2!! I think you would find it better just to leave the denominator as (x+ 1)2.

    As I said, your denominator is wrong. (1+1)2= 4.

    Yes, see my comments above.





     
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