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Monotony of a hyperbolic function.

  1. Apr 5, 2007 #1
    i am having some difficulties in proving that

    y=chx=(e^x+e^(-x))/2 is decreasing in the interval (- infinity,0) and increasing in (0, infinity)

    i know that a function is increasing in (a,b) if for two variables from that interval let's say x' and x" that are related x'<x" than f(x')<f(x"), sot at this case f is increasing in (a,b).

    if x'<x", but f(x')>f(x"), than f is decreasing in (a,b).
    here is what i did with chx:

    x'<x", x',x">0 , then from the monototy of exponential functions we know that e^x'<e^x" and ( 1/e^x")<(1/e^x')..

    here is where i get stuck..

    any help????
  2. jcsd
  3. Apr 5, 2007 #2


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    Try looking at the slope (derivative)...
  4. Apr 5, 2007 #3
    so you are saying that there is no way to determine wheather it is increasing or decreasing, the way i tackled it???
    thnx indeed
  5. Apr 5, 2007 #4


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    Well, you can do it your way but you'll have to work harder because there's a problem--exactly where you got stuck. Use the knowledge that 1/e^x is smaller than e^x, or more exactly, show that e^x' + 1/e^x' < e^x'' + 1/e^x''.

    It's easier to use the derivative because then you just need to show that f'(x) is positive or negative.
  6. Apr 5, 2007 #5


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    Check that, apart from the factor 1/2, the 2nd derivative is e^x+e^(-x)>0. This means that the function is strictly convex and has a unique minimum satisfying e^x - e^(-x)=0 that is, the minimum is at x=0.[Note that the function has sirictly increasing first derivative and it is 0 at x=0].
    Last edited: Apr 5, 2007
  7. Apr 5, 2007 #6
    yeah, when it came to show that:
    e^x' + 1/e^x' < e^x'' + 1/e^x''

    i got stuck. So, even it is harder, there is a way to prove this no?? Different that of derivatives??
  8. Apr 5, 2007 #7
    using first and second order derivatives, i proved what i was looking for, however, can somebody, help me prove the way i started it??
    i am really curious to see how it goes. or at least just give me some hints on how to tackle e^x' + 1/e^x' < e^x'' + 1/e^x'' this.

    thnx for your help
  9. Apr 5, 2007 #8
    I'm confused why you'd want to waste your time on this. Derivatives exist for a purpose and this is a perfect fit. First derivative set to zero will tell you everywhere on the curve where the function changes directions, or where you have a zero slope. Then you need to know where its increasing or decreasing. For this you can plug in numbers and test, or you can just take a second derivative and determine if its a convex up or down which directly tells you whether you are increasing or decreasing.

    Doing it your original way you are left to the task of plotting points and doing everything manually. Without calculus I do not think you can solve that...

  10. Apr 6, 2007 #9
    I did prove that, using derivatives, i was just curious to know if there is a way out, withut plotting points, but in an analytical way to prove the way I originally posted it, on the top.

    thnx for your help.
  11. Apr 6, 2007 #10


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    The proof using derivatives actually gives the proof in your wanted way too when x' and x'' (as denoted by you) are very very close.
  12. Apr 6, 2007 #11


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    BTW, people and situations can be monotonous. The word for functions is monotonicity. (Not a pretty word, but no one asked me when they made it...)
  13. Apr 7, 2007 #12
    thnx, i thought both worked for the same puporse. Now i see i was mistaken!!
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