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Calculus I

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Hey guys. I tried solving these on my own, but I cannot seem to find out how to do this.

    23. What are the steps to finding

    sqrt(3+6*x^2)/(2+5*x) as x goes to infinity?


    41. Find the Horizontal Asymptotes for

    17x/(x^4+1)^1/4 . I'm pretty sure its 17/1, but for some reason, my teacher asks for two, and I thought rational functions only have one horizontal asymptote...



    62. f(x)= 4x^3+13x^2+11x+24 / x+3 when x<-3
    f(x)= 3x^2+3x+A when -3 less than or equal to x

    What is A in order for it to be continuous at -3?


    79. f is continous at (-inf, + inf)

    f(y) = cy+3 range is (-inf,3)
    f(y) = cy^2-3 range is (3,+inf)

    what is C?


    81. Let f(x) = {2x^2+3 x -65) / (x-5)

    Show that f(x) has a removable discontinuity at x=5 and determine what value for f(5) would make f(x) continuous at x=5.
    Must define f(5)=



    Thank you so much in advance >< I've been doing these questions for 3 hours ... I need help to prepare for my calculus test next week !!
     
  2. jcsd
  3. Sep 27, 2008 #2

    HallsofIvy

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    If you tried, what did you do? It is always better to show what you have tried. One thing you can do is say "if x is extremely large, we can ignore "3" in comparison with 6x2 and "2" in comparison to 5x. What do you get if you just drop the 3 and 2?

    A little more rigorous is to divide both numerator and denominator by x. While "going to infinity" is a nuisance, going to 0 is easy- and when x goes to infinity, 1/x goes to 0.


    No, that is not necessarily so. "Horizontal asymptotes" occur is the graph approaches a horizontal line as x goes to infinity or negative infinity. What happens if x is a large negative number? For example, what is the value if x= -100000000?



    What is the definition of continuous? In particular, what are the limits as x goes to -3 from above and below?


    This makes no sense. First you must mean "on" (-inf, +inf), not "at". Second, do you mean "domain" rather than "range"? And, finally, the only way that would make sense is if the two domains are (-inf, 3] and [3,+inf). Assuming those are correct, what are the limits as x goes to 3 from above and below?


    You must define f(5) to be the limit as x approaches 5. f does not have a value at x= 5, given by that formula, because the denominator is 0 at x= 5. But in order to have a limit at x= 5, the numerator must also be ____. And what does that tell you about factoring 2x2+ 3x- 65?



     
    Last edited: Sep 27, 2008
  4. Sep 27, 2008 #3
    HallsofIvy, thanks for replying. I will try to post up what I did here.

    23. What are the steps to finding

    sqrt(3+6*x^2)/(2+5*x) as x goes to infinity?

    i figured out the answer is sqrt(2) x sqrt (3) / 5 I dont get what I'm supposed to do to the numerator with the square roots.

    What I did is turn that function disregarding the +2 and +3 so I got sqrt 6x^2 / 5x ... and I'm not suer what I do next...


    41. Find the Horizontal Asymptotes for

    17x/(x^4+1)^1/4 . I'm pretty sure its 17/1, but for some reason, my teacher asks for two, and I thought rational functions only have one horizontal asymptote...[/quote]
    No, that is not necessarily so. "Horizontal asymptotes" occur is the graph approaches a horizontal line as x goes to infinity or negative infinity. What happens if x is a large negative number? For example, what is the value if x= -100000000?

    What I did first is fix the denominator and make it become 17x / x+1 because I multiply the roots. I disregard the +1 and get 17x/x ... This has two horizontal asymptotes?
    I thought because the ratio of the coefficient of 17x and x is 1 , n=m=1 therefore it is 17/1 ? please correct me if i'm wrong . if x is -10000 then it is neg infinity .



    62. f(x)= 4x^3+13x^2+11x+24 / x+3 when x<-3
    f(x)= 3x^2+3x+A when -3 less than or equal to x

    What is A in order for it to be continuous at -3?[/quote]
    What is the definition of continuous? In particular, what are the limits as x goes to -3 from above and below?

    continuous means there are no breaks , basically being able to draw it on a graph without lifting up pencil. also there is no discontinuity. as x goes to -3 when x<-3 it goes to 0?
    I still dont understand what A has to be...


    79. f is continous at (-inf, + inf)

    f(y) = cy+3 range is (-inf,3)
    f(y) = cy^2-3 range is (3,+inf)

    what is C?

    This makes no sense. First you must mean "on" (-inf, +inf), not "at". Second, do you mean "domain" rather than "range"? And, finally, the only way that would make sense is if the two domains are (-inf, 3] and [3,+inf). Assuming those are correct, what are the limits as x goes to 3 from above and below?

    Sorry I did mean on, and yes it is the domain. the limits as x goes to 3 from above is -inf ?





    81. Let f(x) = {2x^2+3 x -65) / (x-5)

    Show that f(x) has a removable discontinuity at x=5 and determine what value for f(5) would make f(x) continuous at x=5.
    Must define f(5)=

    You must define f(5) to be the limit as x approaches 5. f does not have a value at x= 5, given by that formula, because the denominator is 0 at x= 5. But in order to have a limit at x= 5, the numerator must also be ____. And what does that tell you about factoring 2x2+ 3x- 65?

    factoring 2x^2+3x-65 would give me (2x+13)(x-5) so i'll cancel out and get (2x+13). So do I use direct substitution? f(5)= 2(5)+13 = 23?


    Thanks for all the help . I appreciate it. I tried to make the colors different so it won't seem too confusing.
     
  5. Sep 27, 2008 #4

    HallsofIvy

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    Well, what is [itex]\sqrt{6x^2}/5x[/itex] for x positive?

    To see that you can "disregard" the +2 and +3, divide both numerator and denominator by x. Since x is going to infinity it is not 0 so we certainly can do that and see that [itex]\sqrt{2+ 6x^2}/(2+5x)= \sqrt{2/x^2+ 6}/(2/x+ 5)[/itex]. Now, as I said before, as x goes to infinity, 1/x goes to 0.


    No, that is not necessarily so. "Horizontal asymptotes" occur is the graph approaches a horizontal line as x goes to infinity or negative infinity. What happens if x is a large negative number? For example, what is the value if x= -100000000?

    What I did first is fix the denominator and make it become 17x / x+1 because I multiply the roots. I disregard the +1 and get 17x/x ... This has two horizontal asymptotes?
    I thought because the ratio of the coefficient of 17x and x is 1 , n=m=1 therefore it is 17/1 ? please correct me if i'm wrong . if x is -10000 then it is neg infinity .

    What happened to the fourth power and 1/4 power? (x4+ 1)1/4 is NOT equal to x+1. In fact, even (x4)1/4 is not equal to x- it is equal to |x|. Do you see why you get two different asymptotes?

    And, why in the world would you think doing any calculation (where the denominator is not 0) would give "infinity"? If x= -10000, 17x/(x4+1)1/4 is -170000/10000.00000000000025= -16.999999999999999575, not anywhere near "infinity"!

    Is there any reason you keep writing "17/1" rather than "17"?


    What is the definition of continuous? In particular, what are the limits as x goes to -3 from above and below?

    No, that is not the definition of continuity, it is a general property. The definition of "continuity of f(x) at x= a" requires three things: 1) f(a) exists; 2) [itex]\lim_{x\rightarrow a} f(x)[/itex] exists; 3) those are equal: [itex]f(a)= \lim_{x\rightarrow a} f(x)[/itex].

    Of course, the limit exists if the two "one sided limits" exist and are the same. Often you can find the limit of f(x), as x goes to a, by finding f(a) because, often, the functions we work with are continous. If you try finding the limit of f(x) as x approaches -3 "from below" by evaluating (4x3+13x2+11x+24) /(x+3) by evaluating at x= -3, you get 0/0 so that doesn't work. But the fact that the numerator is 0 at x= -3 tells us that x-(-3)= x+ 3 is a factor. Find the other factor so you can cancel the "x+ 3" terms and find the limit.

    The limit from above, [itex]lim_{x\rightarrow -3^+} 3x^2+ 3x+ A[/itex] is easy: it is a polynomial, so continuous so the limit is 3(-3)2+ 3(-3)+ A= 27- 9+ A= 18+ A.

    In order that the given function be continuous at x= -3. Those two limits must be the same. Set them equal and solve for A.

    Both "parts" of this function are polynomials and so the limits "from below" and "from above" can be found by evaluating them. What is c(3)+ 3 and c(32)- 3? Set them equal and solve for c.




    Yes, that's right. (2x+13)(x-5)/(x-5)= 2x+13 as long as x is not 5, and the limit as x approaches 5 depends only on what the value is for x close to 5, not equal to 5. Since (2x2+ 3x- 65)/(x-5)= 2x+ 13 for all x other than 5, they have the same limit at 5. and since 2x+ 13 is a polynomial, it is continuous and its limit is its value at x= 5.
     
  6. Sep 27, 2008 #5
    41. horizontal asymptotes are 17 and -17 then ?
    62. i still dont really understand... what do you mean by x-(-3)= x+ 3 is a factor? i dont get the 4x^3+13x^2+11x+24/ x+3 when x<-3 part but I understand the other
    81. the answer is 23 then?
    79. the answer is 1 ?

    another question is:
    the vertical asymptote of -x^3/x-2 is x=2 right ?

    sorry this is my first year learning calculus and im not that familiar with the rules yet
     
  7. Sep 28, 2008 #6

    HallsofIvy

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    Yes, the horizotal asymptotes are 17 and -17.

    If p(x) is a polynomial and p(a)= 0, then (x- a) is a factor of p(x): p(x)= (x-a)q(x) where q(x) is a polynomial of lower degree. In this case, since 4(-3)3+ 13(-3)2+ 11(-3)+ 23= 0, 4x3+ 13x2+ 11x+ 24= (x+3)q(x). You can find q(x) by dividing 4x3+ 13x2+ 11x+ 24 by x+3 (it was for precisely this purpose that "synthetic division" was developed). To put you out of your misery, 4x3+ 13x2+ 11x+ 24= (x+3)(4x2+ x+ 8). What is (4x3+ 13x2+ 11x+ 24)/(x+3)?

    There are two parts to the question: the first is to show that there is a removeable discontinuity. You can do that by showing that the limit of the function, as x goes to 5, exists. Then the answer to the second part is that limit which is, yes, 23.

    Yes, 9c- 3= 6c+ 3 gives c= 1.


    Yes, that fraction is not defined at x= 2 so the graph cannot cross that vertical line. That is a vertical asymptote. (Notice that the graph cannot cross a vertical asymptote but can cross a horizontal asymptote. That is the because the definition of "function" does not allow two (x,y) points on the graph to have the same x value, but does allow two points with the same y value.)

     
  8. Sep 28, 2008 #7
    thanks i understand now . sorry for the trouble
     
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