Solving Functions without a Calculator: A Comparison of A and B on [0;1/10]

  • Thread starter mohlam12
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In summary, the problem asks which function is bigger, A or B. The function with a constant term, 1, in the numerator is greater than the function without the constant term.
  • #1
mohlam12
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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 going to 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 :)
 
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  • #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.
 
  • #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 !?
Thanks,
 
  • #4
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.
 
  • #5
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:
  • #6
"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.
 
  • #7
Okay !
I solved it and I appreciate your help! :)
 

1. What is a function?

A function is a block of code that performs a specific task and can be called multiple times throughout a program. It takes in inputs, performs operations on them, and returns an output.

2. What are the main components of a function?

A function has three main components: the function name, the parameters (inputs) that it takes in, and the code block that contains the instructions for the function to execute.

3. How do I call a function?

To call a function, you use its name followed by parentheses. If the function has parameters, you would include the values for those parameters inside the parentheses.

4. Can a function have multiple return statements?

Yes, a function can have multiple return statements. However, only one of them will be executed each time the function is called. Once a return statement is reached, the function will stop executing and return the value specified in that return statement.

5. Can a function call itself?

Yes, a function can call itself, this is known as recursion. It is a useful technique for solving certain types of problems, but it is important to have a base case or stopping condition in order to prevent infinite recursion.

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