1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Continuity of a Function.

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Find all values of the parameter a>0 such that the function

    [tex]f(x)=\left\{\begin{array}{cc}\frac{a^x+a^{-x}-2}{x^2},x>0\\3ln(a-x)-2,x\leq0\end{array}\right[/tex]



    3. The attempt at a solution

    [tex]\lim_{x\rightarrow 0}\frac{a^x+a^{-x}-2}{x^2}=0[/tex]

    [tex]0=3ln(a-0)-2\rightarrow a=e^{2/3}[/tex]
     
  2. jcsd
  3. Oct 21, 2007 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Why should the limit value be 0?? :confused:

    All that is required is that AT x=0, both expressions should attain the same value!

    Now, consider the function's expression for the non-positives.
    Clearly, its limit at x=0 is 3ln(a)-2.
    Thus, you are to determine those values of "a" so that the limiting value of the function expression for the positives equals 3ln(a)-2 as well.
     
  4. Oct 21, 2007 #3
    ok
    [tex]\lim_{x\rightarrow 0}\frac{a^x+a^{-x}-2}{x^2}=\frac{ln^2a}{2}[/tex]
    by L'Hopital Rule

    [tex]\frac{ln^2a}{2}=3ln(a)-2[/tex]

    [tex]ln^2a-4lna+4=0[/tex]
     
    Last edited: Oct 21, 2007
  5. Oct 21, 2007 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The limit of the upper expression should be [tex]\ln^{2}(a)[/tex]

    Thus, you have the quadratic in ln(a) to solve:
    [tex]\ln^{2}(a)=3\ln(a)-2[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Continuity of a Function.
  1. Continuous Functions (Replies: 2)

  2. Continuity function (Replies: 8)

Loading...