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Y=2(3^x)-1 domain and range?

  1. Dec 7, 2004 #1
    I was given the equation y=2(3^x)-1 and was told state the domain, range, x-intercept(s) y-intercept(s) and asymptote(s)
    I graphed the equation and I think the x-intercept is -0.63
    I subbed in 0 for the x value on the graphing calculator and got y-intercept=1
    the vertical asymptote is the same as the x-intercept so this =-0.63 and then the horizontal asymptote always equals 0.

    If this is all right then I would like to know how to find out the domain and range? :confused: Can someone please help.
  2. jcsd
  3. Dec 7, 2004 #2
    look into the definition of domain and range
    Domain :: Those values of x for which the function is defined on [tex]\R[/tex]
    Range :: The value function takes on the domain
  4. Dec 7, 2004 #3
    I know the definition domain is all the values of x that satisfy equation, and range is all the values of y that can satisfy the equation, but I dont know how to figure them out, and dont know if the other stuff I did is also right? :redface:
  5. Dec 8, 2004 #4


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    Can you find the range of the function 3^x?
    Can you use this to find the range of 2(3^x)?
    Can you use this to find the range of 2(3^x)-1?
    It might help if you think of the graph of 3^x, and how the graph of your function relates to it.

    The domain should be easier-are there any values of x where your function is undefined?

    You say your x-intercept came from graphing it-can you find the x-intercept algebraicaly? Same with the y-intercept, you should be able to find y when x=0 by hand.
  6. Dec 8, 2004 #5
    This is a fun question. I think you will get the best results from graphing and looking at the table. There is certainly an asymptote for y, and knowing that the domain and range should come quite easily.

    -EDIT: With this question, it is most important to realize your calculator will round the answers in the table.
    Last edited: Dec 8, 2004
  7. Dec 8, 2004 #6


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    It's not really necessary nor even helpful to graph this. The domain is, as you say, "all values of x which satisfy the equation" which means all values of x for which you can perform the operation. Here, y= 2(3x)-1. Are there any values of x for which you can't find 3x? As far as the range is concerned, it is helpful to remember that 3x is never negative.
  8. Dec 8, 2004 #7
    The limit on [tex]2 ( 3^x )[/tex] just becomes more apparent after the equation is graphed if it is not immediately obvious.
  9. Dec 8, 2004 #8
    OK guys I think I got the domain = {x:XER} and the range {YER,y>=1}
    Is this right?
    also can someone show me how to write how to find this out? I guesed looking at my calculator, and the table of values.

    Also is my x-intercept = -0.63?
    and my y-intercept is 1

    um I think there is no vertical asymptote but dont know y? :yuck:
    but the horizontal asymptote is 0?
  10. Dec 8, 2004 #9
    I was under the assumption that an asymptote was a line the function would not cross, and this function does cross the line y=0.

    Incorrect, there is no problem when y is less than 1
  11. Dec 8, 2004 #10
    ok sorry I meant {y: y<=-1,YER} Is that right? There is no vertical asymptote, or horizontal asymptote because this is not a reciprocal function? Is this a good reason? My x-intercept is correct? -0.63? please help me out! :cry:
  12. Dec 9, 2004 #11
    Yes, you are almost right. Your only error was saying y could equal -1. It's
    y > -1. Also, would not the horizontal asymptote be y=-1? And yes, the x-int is roughly equal to -0.63, but were I you, I would come up with n exact value for the x-into algebraically.
  13. Dec 9, 2004 #12
    K i m sooo dumb I m going to find the x intercept algebraically but got stuck at 1/2=3^x how do i get the 3 on the other side?
  14. Dec 9, 2004 #13


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    ONLY THROUGH LOGARITHMS.But if a were u,i'd put my computer to plot "3^x" and find the interception with the horizontal line "y=1/2".
  15. Dec 9, 2004 #14
    which would be log base 1/2 of 3 = x . this means that (log base 10 of 1/2)/(log base 10 of 3) = x or so i think plz correct me.
  16. Dec 9, 2004 #15
    Soo is the horizontal asymptote x=-1? and the vertical x=-0.63? IM NOT SURE :cry:

    PLEASE CAN SOMEONE HELP ME OUT????????? :confused:
    Last edited: Dec 9, 2004
  17. Dec 9, 2004 #16

    Tom Mattson

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    Yes. If you take the limit as x goes to negative infininity, you get y=-1.

    There is no vertical asymptote. A vertical asymptote occurs at an infinite discontinuity, but the exponential function is continuous for all x.
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