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A function's question

  1. Dec 14, 2005 #1
    hi everyone,
    i have to solve the problem below:
    1) which is the bigger number, (without using a calculator)
    B= (0.9999995)^2 / 0.9999998 & A=1.0000004/(1.0000006)^2
    Then, compare those two functions on the intervall of [0;1/10] (without using a calculator)
    f(x)=(1-5x)² / 1-2x & g(x)=1+4x / (1+6x)²
    so, first of all, i noticed that if u put 10^-7 on f(x) you'll have the number A, and if you put 10^-7 on g(x) you'll get the number B
    therefore we'll have to compare the two functions to see which is bigger, A or B. But i dun know how i am gonna compare them (do not use a calculator!!)
    i tought about subsituing f(x) from g(x), here is what i got:
    f(x)-g(x) <=> 51x²+60x^3+900x^4 / 1+10x+12x²+72x^3
    so i have to see if this is bigger/smaller than 0.
    ummm, i am not sure, but we have 0<x<1/10
    so 0<51x²<51/100 & 0<60x^3<3/50 & 0<900x^4<9/100
    therefore 0<51x²+60x^3+900x^4<63/100
    and for the denominator we have
    0<10x<1 & 0<12x²<3/25 & 0<72x^3<9
    therefore 0<1+10x+12x²+72x^3<278/25
    and finally, a positive number over a positive number is a positive number
    so f(x)>g(x) <=> A>B (because f(x) and g(x) are increasin over [0;1/10] << I think :s )
    so, is all this right ?! :uhh:
    thanks for ur help :)
  2. jcsd
  3. Dec 14, 2005 #2
    Well, as for the first thing (the one with "dont use a calculator"), imho you should be able to do that with your head only.
  4. Dec 14, 2005 #3
    Okay, but they asked us to compare the two functions as to use them to see if A>B or A<B.
    But anyhow, I am wondering if my calculations and the method I used is right !?
  5. Dec 14, 2005 #4


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    It's probably easier to think about one function divided by the other, instead of subtracted. Then, when deciding which one is bigger, you see if this quotient is greater than or less than one at 10^-7. The easiest way to do that is to ignore all the x^2 and higher terms, since they will look like 10^-14, etc, and so will be insignificant compared to the terms with only the first power of x. For example, if you ended up with (1+2x+4x^2)/(1+x+56x^2), this would be approximately (1+2x)/(1+x) for very small x, and so greater than one.
  6. Dec 14, 2005 #5


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    For A vs. B I would have let [itex]1-\epsilon = 0.9999995[/itex] and [itex]1+\delta = 1.0000004[/itex], etc. then try to simplify. (Here [itex]\epsilon \approx \delta[/itex] but one can also define [itex]\epsilon = \delta 5/4[/itex].)
    Last edited: Dec 14, 2005
  7. Dec 15, 2005 #6


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    "f(x)-g(x) <=> 51x²+60x^3+900x^4 / 1+10x+12x²+72x^3"

    IF that were right then the answer would be obvious: all the numbers are positive!

    HOWEVER, it's not right. For one thing there is a constant term, 1, in the numerator, and clearly the leading coefficient in the denominator is -72, not + 72.

    If you do what StatusX suggested, divide one function by the other, and think about what happens if x is very small- assuming one of the functions is larger than the other for all x in 0 to 1/10- it should be straightforward.
  8. Dec 15, 2005 #7
    Okay !!
    I solved it and I appreciate your help! :)
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