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Various Problems for Precalculus Exam, Unit 1

  1. Jan 1, 2009 #1

    jacksonpeeble

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    1. The problem statement, all variables and given/known data
    6. Draw a graph of the function f, and determine the intervals on which f is increasing and on which f is decreasing where f(x)=2x3-3x2-12x

    9. Suppose the graph of f is given. Using transformations, describe how the graph of the following function can be obtained from the graph of f: y=2f(x+4)-5

    21. Find the equation of the horizontal asymptote of the function g(x)=[tex]\frac{x^{3}}{3x^{3}-2x}[/tex].

    2. Relevant equations
    6. f(x)=2x3-3x2-12x

    9. y=2f(x+4)-5

    21. g(x)=[tex]\frac{x^{3}}{3x^{3}-2x}[/tex]

    3. The attempt at a solution
    6. Calculated graph in which local maximum was (-1,7) and local minimum was (2,-20). Said that for -[tex]\infty[/tex] through -1, y is increasing and for -1 through 2, y is decreasing, and for 2 through [tex]\infty[/tex], y is increasing.

    9. (x+4) means left 4, -5 means down 5, 2 means twice the height, - means reflect over x-axis.

    21. g(x)=[tex]\frac{x^{2}}{3x^{2-}2}[/tex] so the horizontal asymptote must equal 0 because the lim g(x) as x[tex]\rightarrow\infty[/tex]=0.
     
    Last edited: Jan 1, 2009
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  3. Jan 1, 2009 #2
    6. I think it was cubic (you have square)
    looks good
    see: http://img515.imageshack.us/img515/498/graphmr5.jpg

    9. x+4 shifts to left, -5 down.. don't know about "3 means down 3, - means reflect over x-axis." don't see them in the question.

    21. Looks wrong.. trying factoring out x^2 from the denominator ...
     
  4. Jan 1, 2009 #3

    jgens

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    For 21, try multiplying and dividing by x^3 and then find the limit.

    You could also go about it by reasoning that as x gets arbitrarily large 3x^2 - 2 -> 3x^2. The limit is easy to find then.
     
  5. Jan 1, 2009 #4

    jacksonpeeble

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    Thanks! What did you use to generate that graph? (You're correct about the cube rather than square.)

    For #9, I was simply trying to find the significance of the other numbers (and that's what I was confused about). Are you saying that all I needed to find were the two that you already said were correct?

    As for #21, I'm just stumped. (Also, thank you to jgens for your response, but I'm afraid I don't know what you mean.) Could somebody please elaborate some more and possibly break the procedure down for me? You can't factor out x^2 from the denominator, can you? The second term doesn't have x^2 in it...
     
  6. Jan 1, 2009 #5
    #
    http://www.padowan.dk/graph/

    #
    It's just this thing. You talked about 4 and -5, but there isn't any 3 ...
    y=2f(x+4)-5

    #21
    Yes, you can factor out. If you don't see how, multiply the denominator by x^2/x^2.

    x^2*[1/x^2 * den] and you have your x^2 out.
     
  7. Jan 1, 2009 #6

    jacksonpeeble

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    Thanks! I see what you mean now! I really appreciate the help.

    About number 9 - I accidentally copied part of the next question on our review sheet into this post. I changed it. It was supposed to say "2 means 2x as high." The negative means flip of axis was accurate. Is this updated version correct :wink:
     
  8. Jan 1, 2009 #7
    For questions like #9, it might be helpful to pick a function.... say f(x) = x^2 or whatever you want and then see what's happening i.e. what's the difference between x^2 and (x+4)^2 (i.e. f(x) and f(x+4))?
     
  9. Jan 1, 2009 #8

    jgens

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    Sorry my post was difficult to understand, I'll explain it better.

    If we multiply and divide the function x^3/(3x^3-2x) by x^3 we get (with a little algebraic manipulation) 1/(3-2/x^2). The limit is easy to find then.

    My second suggestion arises from the fact that for large x, 3x^2 - 2 is essentially equivalent to 3x^2 - you can show this by using any large value for x; eventually, the difference becomes negligible. Using that logic, it's easy to find the limit.
     
  10. Jan 1, 2009 #9
    I remember reading from an Algebra 2 book this rule to find horizontal asymptotes, which is essentially what jgens et. al. said:
    Both the numerator and denominator are polynomials. Compare the degrees of these polynomials. (To save typing, d_num = "degree of the polynomial in the numerator," and d_den = "degree of the polynomial in the denominator.)
    1) If d_num < d_den, then the horizontal asymptote is y = 0.
    2) If d_num = d_den, then the horizontal asymptote is y = the quotient of the leading coefficients (**** this is what you should use for your #21)
    3) If d_num > d_den, then there is no horizontal asymptote. (There is a slant asymptote.)


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