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Math Help - PLEASE help with rational functions assignment

  1. #1
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    PLEASE help with rational functions assignment

    I have this assignment due wednesday.. and I've started it.. but I want to make sure I'm on the right track.. and some of them I have no clue where to start. Anyways.. if anyone can help it would be much appreciated

    1. Determine the derivative of
    f(x)= 2x^2 - 1
    .............x
    from first principles. Use the quotient rule to verify your answer.

    2. The graph of
    f(x)= ax + b
    ......(x -1)(x-4)
    has a horizontal tangent line at (2,-1). Determine the values of a and b.

    3. The tangent line to the curve defined by
    g(x)= x^4 - 2x^3 +3
    ..............-6x
    at x= -1 intersects the curve at two other points P and Q. Determine the coordinates of P and Q algebraically. Illustrate this situation with a graph. (to graph g you need to create a table of values)
    HINT: You need to research L'hopital's rule in order to complete this question



    I have no clue what L'hopital's rule is, because we've never learned it, so if someone could also explain that to me.. that would be great.

    Thank-you!
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  2. #2
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    Quote Originally Posted by jmailloux View Post
    .. if anyone can help it would be much appreciated

    1. Determine the derivative of
    f(x)= 2x^2 - 1
    .............x
    from first principles. Use the quotient rule to verify your answer.

    ...
    Hello,

    f(x)=\frac{2x^2-1}{x}=2x-\frac{1}{x}. Now calculate the derivative:
    f'(x)=2+\frac{1}{x^2}. Using quotient rule:
    f'(x)=\frac{x\cdot (2\cdot 2x)-(2x^2-1)\cdot 1}{x^2}=\frac{4x^2-2x^2+1}{x^2}=\frac{2x^2+1}{x^2}=2+\frac{1}{x^2}

    EB
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  3. #3
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    Quote Originally Posted by jmailloux View Post
    .. if anyone can help it would be much appreciated

    2. The graph of
    f(x)= ax + b
    ......(x -1)(x-4)
    has a horizontal tangent line at (2,-1). Determine the values of a and b.
    ...
    Hello,

    you have the function:
    f(x)=\frac{ax+b}{(x-1)(x-4)}. And you know f(2) = -1 and f'(2) = 0:

    f(2)=\frac{2a+b}{(2-1)(2-4)}=-1\ \Longleftrightarrow\ 2a+b=2

    f'(x)=\frac{(x-1)(x-4)\cdot a-(ax+b)\cdot (2x-5)}{((x-1)(x-4))^2}. Don't expand that. Plug in x = 2. Then f'(2) = 0. Multiply by the denominator :

    -2a+2a+b = 0. Therefore b = 0 and a = 1.

    Your function becomes:

    f(x)=\frac{x}{(x-1)(x-4)}

    EB
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  4. #4
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    Quote Originally Posted by jmailloux View Post
    ... if anyone can help it would be much appreciated
    ...
    3. The tangent line to the curve defined by
    g(x)= x^4 - 2x^3 +3
    ..............-6x
    at x= -1 intersects the curve at two other points P and Q. Determine the coordinates of P and Q algebraically. Illustrate this situation with a graph. (to graph g you need to create a table of values)
    HINT: You need to research L'hopital's rule in order to complete this question


    I have no clue what L'hopital's rule is, because we've never learned it, so if someone could also explain that to me.. that would be great.
    Thank-you!
    Hello,

    I'll show you a way where you don't have to use rule de l'Hopital. (There is actually no need to use this rule!)

    I'll use: g(x)=\frac{x^4-2x^3+3}{-6x}=-\frac{1}{6}x^3+\frac{1}{3}x^2-\frac{1}{2x},\ x\neq0

    Then g(-1) = 1

    g'(x)=-\frac{1}{2}x^2+\frac{2}{3}x+\frac{1}{2x^2}
    Thus g'(-1)=-\frac{2}{3}

    The tangent t becomes: y= t(x)=-\frac{2}{3}x+\frac{1}{3}

    Calculating the intercept mean: g(x) = t(x):
    -\frac{1}{6}x^3+\frac{1}{3}x^2-\frac{1}{2x}=-\frac{2}{3}x+\frac{1}{3}
    Multiply by (-6x):
    x^4-2x^3-4x^2+2x+3=0. You already know that x = -1 must be a solution of this equation. So divide the LHS of the equation by (x+1) and you'll get:
    x^3-3x^2-x+3=0
    x^2(x-3)-(x-3)=0
    (x^2-1)(x-3)=(x+1)(x-1)(x-3)0.

    So you get the additional solutions: x = 1 and x = 3.

    I'll leave the rest for you. Good luck!

    EB
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  5. #5
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    Quote Originally Posted by jmailloux View Post
    ... if anyone can help it would be much appreciated
    ...
    3. The tangent line to the curve defined by
    g(x)= x^4 - 2x^3 +3
    ..............-6x
    ...
    Hello again,

    for confirmation only I send you the graph of the function and the tangent line.

    EB
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