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Math Help - function question

  1. #1
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    Question function question

    This is the function:
    f(x)=(x^2-Bx+4)/(x-A)

    given:
    x=2 critical point
    y=x+5 is an oblique asymptote

    I need to find A and B and decide if x=2 is min or max.

    i got to f'(x)= (x^2-2Ax-4+BA)/{(x-A)^2}
    but I'm not even sure that I need a derivative here.

    can someone explain what would be the way to solve this?
    Please, if you can, show your work so that I can understand it.

    THANKS!
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  2. #2
    MHF Contributor red_dog's Avatar
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    Indeed, the derivative is \displaystyle f'(x)=\frac{x^2-2Ax+AB-4}{(x-A)^2}.
    If x is a critical point then f'(2)=0\Rightarrow A(B-4)=0. (1)

    The oblique asymptote is y=mx+n where \displaystyle m=\lim_{x\to\infty}\frac{f(x)}{x} and \displaystyle n=\lim_{x\to\infty}[f(x)-mx].

    Now, \displaystyle m=\lim_{x\to\infty}\frac{x^2-BX+4}{x^2-Ax}=1.
    \displaystyle n=\lim_{x\to\infty}\left(\frac{x^2-Bx+4}{x-A}-x\right)=\lim_{x\to\infty}\frac{(A-B)x+4}{x-A}=A-B=5 (2)

    From (1) and (2) we get:
    I) A=0,B=-5\Rightarrow x=2 point of minimum
    II) A=9,B=4\Rightarrow x=2 point of maximum.
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  3. #3
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    Thanks for your reply!
    I still don't understand the second part.
    How did you decide what A goes with what B
    and how do you know what's max and whats min?
    If you could clear this up for me that would be great
    thanks!
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  4. #4
    MHF Contributor red_dog's Avatar
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    For A=0,B=-5\Rightarrow \displaystyle f'(x)=\frac{x^2-4}{x^2}
    The roots of f'(x) are x_1=-2,x_2=2.
    Then f'(x)<0,\forall x\in(0,2), so f is decreasing and
    f'(x)>0,\forall x\in(2,\infty), so f is increasing.
    Then x=2 is a point of minimum.

    For A=9,B=4\Rightarrow \displaystyle f'(x)=\frac{x^2-18x+32}{(x-9)^2}.
    The roots of f'(x) are x_1=2,x_2=16.
    f'(x)>0,\forall x\in(-\infty,2), so f is increasing and
    f'(x)<0,\forall x\in(2,9), so f is decreasing.
    Then x=2 is a point of maximum.
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  5. #5
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    thanks for all your help
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