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Math Help - Function analysis

  1. #1
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    Post Function analysis

     <br />
f(x)=\frac{x^2+3x-4}{x^2+x-2}<br />

    a. Find the domain of function

    b. Find \lim_{x\to1}f(x)

    c. Find all vertical and horizontal asymptotes

    d. Find \lim_{x\to\infty}f(x)


    Here are my answers...are they correct?

    a. All real numbers not equal to -2

    b. \infty

    c. Vertical: y=1, -1
    Horizontal: x=2

    d. \lim_{x\to\infty}f(x)=1
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  2. #2
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    Quote Originally Posted by live_laugh_luv27 View Post
     <br />
f(x)=\frac{x^2+3x-4}{x^2+x-2}<br />

    a. Find the domain of function

    b. Find \lim_{x\to1}f(x)

    c. Find all vertical and horizontal asymptotes

    d. Find \lim_{x\to\infty}f(x)


    Here are my answers...are they correct?

    a. All real numbers not equal to -2

    b. \infty

    c. Vertical: y=1, -1
    Horizontal: x=2

    d. \lim_{x\to\infty}f(x)=1

    f(x)=\frac{x^2+3x-4}{x^2+x-2}

    a) Only half right.

    Polynomials are continuous everywhere, so the quotient of polynomials is continuous everywhere except where the denominator is 0.

    So if the denominator is 0, we have

    x^2 + x - 2 = 0

    (x + 2)(x - 1) = 0

    x = -2 or x = 1.


    Thus the domain of the function is \mathbf{R}\backslash\{-2, 1\}.


    b) Incorrect.

    Notice that we can rewrite the function as

    f(x) = \frac{(x + 4)(x - 1)}{(x + 2)(x - 1)}

     = \frac{x + 4}{x + 2}.


    Now find \lim_{x \to 1}f(x).


    c) Nearly correct.

    We can further rewrite the function as

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


    There will be a vertical asymptote at x = -2 and a horizontal asymptote at f(x) = 1.


    d) Correct.

    As x \to \infty, x + 2 \to \infty.

    Therefore \frac{2}{x + 2} \to 0.

    Thus 1 + \frac{2}{x + 2} \to 1.
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  3. #3
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    Post

    So for b, would the correct answer be -2? I'm still unsure of this part.
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  4. #4
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    Quote Originally Posted by live_laugh_luv27 View Post
    So for b, would the correct answer be -2? I'm still unsure of this part.
    No.

    \lim_{x \to 1}\frac{x + 4}{x + 2} = \frac{1 + 4}{1 + 2}

     = \frac{5}{3}.
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