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Math Help - calculus graph, please check, thanx

  1. #1
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    calculus graph, please check, thanx

    Qu.

    Use graph sketching strategy to sketch the graph of the function f(x) = 4x -15/ x^2 -9

    Step 1 The domain of f is all natural numbers except + 3 or - 3 when x^2 - 9 = 0

    Step 2 Function f in nether even or odd

    step 3 x intercept , 4x -15 = 0, x = 3 3/4
    y intercept f (0) = 5/3

    step 4

    f(x) = 4x -15/ (x-3) (x+3)


    f is positive (3 3/4, infinite), and negative on (-infinite, -3), (-3,3), and (3, 3 3/4)

    Step 5

    the derivative of f(x) = -4x^2 -36 +30x/ (x^2-9)^2

    Then I got stuck....please help
    Last edited by fair_lady0072002; July 13th 2006 at 09:02 AM.
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  2. #2
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    Are you wanting the derivative of:

    \frac{-4x^{2}+30x-36}{(x^{2}-9)^{2}}?.

    Rewrite as such and use the product rule:

    (-4x^{2}+30x-36)(x^{2}-9)^{-2}

    (-4x^{2}+30x-36)(-2)(x^{2}-9)^{-3}(2x)+(x^{2}-9)^{-2}(-8x+30)

    \frac{16x^{3}-120x^{2}+144x}{(x^{2}-9)^{3}}+\frac{-8x+30}{(x^{2}-9)^{2}}

    Multiply the top and bottom of the right side by (x^{2}-9)

    and simplify:

    \Large\frac{2(4x^{3}-45x^{2}+108x-135)}{(x^{2}-9)^{3}}
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  3. #3
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    Quote Originally Posted by fair_lady0072002
    Qu.

    Use graph sketching strategy to sketch the graph of the function f(x) = 4x -15/ x^2 -9

    Step 1 The domain of f is all natural numbers except + 3 or - 3 when x^2 - 9 = 0

    Step 2 Function f in nether even or odd

    step 3 x intercept , 4x -15 = 0, x = 3 3/4
    y intercept f (0) = 5/3

    step 4

    f(x) = 4x -15/ (x-3) (x+3)


    f is positive (3 3/4, infinite), and negative on (-infinite, -3), (-3,3), and (3, 3 3/4)

    Step 5

    the derivative of f(x) = -4x^2 -36 +30x/ (x^2-9)^2

    Then I got stuck....please help
    Sketching curves has always been one of my favorite thing in calculus. I will you show you my way.
    -------------------------------------------
    Step 1: Find domain
    Find the domain of the function,
    f(x)=\frac{4x-15}{x^2-9}
    Note, x^2-9\not = 0 thus, x\not = \pm 3

    Step 2: Symettry
    Does f(-x)=f(x)
    You can see that,
    \frac{4(-x)-15}{(-x)^2-9}=\frac{-4x-15}{x^2-9}--->No
    Does f(-x)=-f(x)
    You can see that,
    \frac{4(-x)-15}{x^2-9}=\frac{-4x-15}{x^2-9} and that,
    -f(x)=\frac{-4x+15}{x^2-9}---> No

    Step 3: Asymptotes
    Yes it does at x=-3,3 (since numerator is non-zero while denominator is zero).

    I countinue later, it takes longgg time.
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  4. #4
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    ThePerfectHacker- thanks, continued

    Horizontal asymptote

    y = 0 ?

    When m < n, the horizontal asymptote is the line y = 0

    where m is the degree of polynomial in the numerator and n is the degree of polynomial in the denominator.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    Sketching curves has always been one of my favorite thing in calculus. I will you show you my way.
    -------------------------------------------
    Step 1: Find domain
    Find the domain of the function,
    f(x)=\frac{4x-15}{x^2-9}
    Note, x^2-9\not = 0 thus, x\not = \pm 3

    Step 2: Symettry
    Does f(-x)=f(x)
    You can see that,
    \frac{4(-x)-15}{(-x)^2-9}=\frac{-4x-15}{x^2-9}--->No
    Does f(-x)=-f(x)
    You can see that,
    \frac{4(-x)-15}{x^2-9}=\frac{-4x-15}{x^2-9} and that,
    -f(x)=\frac{-4x+15}{x^2-9}---> No

    Step 3: Asymptotes
    Yes it does at x=-3,3 (since numerator is non-zero while denominator is zero).

    I countinue later, it takes longgg time.
    I am really stuck, please help
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  6. #6
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    Quote Originally Posted by fair_lady0072002
    I am really stuck, please help
    With what?
    Tell me extacly you were able to do (post it I will check it) and then procede to tell me what you do not know how to solve.
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  7. #7
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    ThePerfectHacker - Graph of a function

    Found derivative of f

    = (-4x +6)(x-6)/(x^2-9)^2

    Problems establishing when f is increasing or decreasing and if any stationary points
    Last edited by fair_lady0072002; July 24th 2006 at 03:39 PM.
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  8. #8
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    Quote Originally Posted by fair_lady0072002
    Found derivative of f

    = (-4x +6)(x-6)/(x^2-9)^2

    Problems establishing when f is increasing or decreasing and if any stationary points
    Fair Lady Check out the thread below

    http://www.mathhelpforum.com/math-he...ead.php?t=3605
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  9. #9
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    Quote Originally Posted by fair_lady0072002
    Found derivative of f

    = (-4x +6)(x-6)/(x^2-9)^2

    Problems establishing when f is increasing or decreasing and if any stationary points
    Multiply the numerator (open parantheses),
    \frac{-4x^2+30x-36}{(x^2-9)^2}
    Use quotient rule,
    \left( \frac{u}{v} \right) = \frac{u'v-uv'}{v^2}
    Thus,
    \frac{(-4x^2+30x-36)'[(x^2-9)^2]-(-4x^2+30x-36)[(x^2-9)^2]'}{(x^2-9)^4} (1)
    Note that,
    (-4x^2+30x-36)'=-8x+30
    And,
    [(x^2-9)^2]'=2(2x)(x^2-9)=4x(x^2-9)--->Rule of chains.

    Now rewrite expression (1) without derivatives,
    \frac{(-8x+30)(x^2-9)^2-4x(-4x^2+30x-36)(x^2-9)}{(x^2-9)^4}
    Factor, that the term (x^2-9) thus,
    \frac{(x^2-9)[(-8x+30)(x^2-9)-4x(-4x^2+30x-36)]}{(x^2-9)^4}
    Cancel,
    \frac{(-8x+30)(x^2-9)-4x(-4x^2+30x-36)}{(x^2-9)^3}
    Open ye olde' parantheses,
    \frac{-8x^3+30x^2+72x-270+16x^3-120x^2+144x}{(x^2-9)^3}
    Thus, combine,
    \frac{8x^3-90x^2+192x-270}{(x^2-9)^3}
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  10. #10
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    Quote Originally Posted by ThePerfectHacker
    ...
    Open ye olde' parantheses,
    \frac{-8x^3+30x^2+72x-270+16x^3-120x^2+144x}{(x^2-9)^3}
    Thus, combine,
    \frac{8x^3-90x^2+192x-270}{(x^2-9)^3}

    Objection, Your Honour!

    72 + 144 = 216. Therefore the 2nd derivative should be:

    \frac{8x^3-90x^2+216x-270}{(x^2-9)^3} as galactus has calculated already

    Greetings

    EB
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