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Questions on ln and e^x graphs

  1. May 27, 2015 #1
    Greetings,
    I have some questions about ln(x) and e^x graphs , with figuring out Domain , range and line of asymptote.
    Q1) How can I know if this graph is ln(x) or e^x

    DSC_1066.JPG
    (I thought it was e^x graph since there's no x-axis intercept , however the answer in marking scheme is:
    Domain : xεR , x>-3
    Range : yεR
    Ast. : x=-3

    So it is ln(x) since there is a x-range/value in the domain .(correct me if i am wrong)

    Q2) what's the difference between the two following two graphs ,, and how can I find the domain and range ?

    DSC_1068.JPG

    DSC_1067.JPG

    Its the Modulus function that confuses me and make it hard to get the domain value.

    *No calculations are required from the above question, as it says "from the figure ... "

    Thank you,
     
  2. jcsd
  3. May 27, 2015 #2

    wabbit

    User Avatar
    Gold Member

    Those questions are weird. There is no way to derive the answers from the figures, other than some guessing.

    For Q2 there are some formulas so you can work with that, but the graphs by themselves do not contain the information required, they only make sense as help in understanding the formulas.

    Regarding the modulus, your formulas have the form ## \ln|X| ## where ## X ## is some expression - so you need to ask (a) what is the domain of the logarithm function, (b) what are the values of ## X ## such that its modulus falls in that domain, and (c) since ## X ## is an expression involving ## x ## , what are the corresponding values of ## x ## ?
     
    Last edited: May 27, 2015
  4. May 29, 2015 #3
    For Q1 you can see the limits on ##x\to-\infty,\,x\to0,\,x\to+\infty## and compare with your graph.
     
  5. May 29, 2015 #4

    NascentOxygen

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    Staff: Mentor

    The fact that it says no calculations are required doesn't mean you shouldn't use your calculator to evaluate some expressions where this will help you to establish range or domain of the equations.

    The plot is neither of these, not exactly. But it does match the shape of one. I suggest that on a sheet of graph paper you manually plot the graphs y = ln(x) and y = e× and having done that compare with the plot you are given.
     
  6. May 29, 2015 #5
    Thank you both
     
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