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Math Help - Graphing using DERIVATIVES

  1. #1
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    Graphing using DERIVATIVES

    I learned how to graph derivatives on Friday, but since I havent reviewed it i pretty much forgot EVERYTHING. Could you guys please post the steps to graphing a curve with the derivative correctly. I need to label the intercepts, relative extrema, points of inflection, and the asymptotes.

    Also, could you solve an example problem lets say x^2 + 1 / x^2 - 9?

    THANKS IN ADVANCE!
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  2. #2
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    Quote Originally Posted by PrecalcKid114 View Post
    I learned how to graph derivatives on Friday, but since I havent reviewed it i pretty much forgot EVERYTHING. Could you guys please post the steps to graphing a curve with the derivative correctly. I need to label the intercepts, relative extrema, points of inflection, and the asymptotes.

    Also, could you solve an example problem lets say x^2 + 1 / x^2 - 9?
    Hello,

    I'm not sure what you mean by "graphing a curve with the derivative correctly". So I'm going to show you what you can do as an examination with the given example:

    f(x)=\frac{x^2+1}{x^2-9}

    1. Domain: The denominator must be unequal zero. The denominator is zero at x = -3 or x = 3. Thus the domain is D=\mathbb{R} \setminus \{-3, 3\}

    2. Asympotes: \lim_{|x|\rightarrow \infty}f(x)=\lim_{|x|\rightarrow \infty}\left(\frac{x^2+1}{x^2-9}\right) = \lim_{|x|\rightarrow \infty}\left(\frac{x^2\left(1+\frac{1}{x^2}\right)  }{x^2\left(1-\frac{9}{x^2}\right)}\right) = 1
    Thus you have 3 asymptotes: x = -3 or x = 3 or y = 1

    3. Y-intercept: f(0)=\frac{1}{-9}

    4. Range: f(x)\leq\frac{1}{-9}~\text{ for all x in }~-3 < x < 3 and f(x) > 1~\text{ for  }~|x| > 3

    That means the range is R=\mathbb{R}\setminus \left\{y | -\frac{1}{9} < y < 1\right\}

    5. Zeros f(x) \ne 0. A fraction is zero if the numerator equals zero and the denominator unequals zero. The term x^2+1 will never become zero.

    6. Relative extrema. Calculate first the first derivative of f:

    f'(x) = \frac{(x^2-9)\cdot 2x - (x^2+1) \cdot 2x}{(x^2-9)^2} =\frac{-20x}{(x^2-9)^2}

    f''(x) = \frac{60(x^2+3)}{(x^2-9)^3} . Don't forget to cancel the factor (x-9)

    f'(x) = 0 ==> x = 0

    f''(0) = -\frac{20}{81} that means you have a relative maximum at x = 0

    7. Point of inflection. Condition: f''(x) = 0~~\wedge~~f'''(x) \ne 0 . f''(x) \ne 0 (see my remark to #5) thus there is no point of inflection.

    8. I've attached a sketch of the graph including the asymptotes.
    Attached Thumbnails Attached Thumbnails Graphing using DERIVATIVES-gebr_rat_fkt_simpl.gif  
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