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Function

by knowledgeisatreasure » Thu Jun 25, 2009 7:31 pm
Hi,

This is my first post...my apologies if I m posting under the wrong topic:

I'm stuck with the following problem:

For which of the following functions F is F(x) = F (1-x) for all x?

a. f(x) = 1-x
b. f(x) = 1-x^2
c. f(x) = x^2 - (1-x)^2
d. f(x) = x^2(1-x)^2
e. f(x) = x / (1-x)

Please provide a step by step explanation. I have read other posts on this problem but I still didn't get the answer.

I really appreciate the help.
Last edited by knowledgeisatreasure on Mon Aug 10, 2009 3:45 pm, edited 1 time in total.
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Re: Function

by pops » Thu Jun 25, 2009 10:24 pm
knowledgeisatreasure wrote:Hi,

This is my first post...my apologies if I posting under the wrong topping:

I'm stuck with the following problem:

For which of the following functions F is F(x) = F (1-x) for all x?

a. f(x) = 1-x
b. f(x) = 1-x^2
c. f(x) = x^2 - (1-x)^2
d. f(x) = x^2(1-x)^2
e. f(x) = x / (1-x)

Please provide a step by step explanation. I have read other posts on this problem but I still didn't get the answer.

I really appreciate the help.
I will give 2 ways to solve this.
method 1 (detailed)
a. f(x) = 1-x
f(1-x) = 1- (1-x) = x

b. f(x) = 1-x^2
f(1-x) = 1 - (1-x)^2 = 1 - (1+x^2-2x) = 2x-x^2

c. f(x) = x^2 - (1-x)^2
f(1-x) = (1-x)^2 - (1 - (1-x))^2 = (1-x)^2-x^2 = -f(x)

d. f(x) = x^2*(1-x)^2
f(1-x) = (1-x)^2*(1-(1-x))^2 = (1-x)^2*x^2 = f(x)

e. f(x) = x / (1-x)
f(1-x) = (1-x)/(1-(1-x)) = (1-x)/x

so d satisfies.

method 2. shorter approach.
put x=0 and 1-0=0 in all these functions and check.

a. 1 when x=0 and 0 when x=1
b. 1 when x=0 and 0 when x=1
c. -1 when x=0 and 0 when x=1
d. 0 when x=0 and 0 when x=1

so we got the answer d.. cant check with 2 so put some other value shows e is not correct.

I find to put values and check is easier for problems on functions.
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by electrico » Fri Jun 26, 2009 12:39 am
welcome to the forum :D

It has already been discuswsed here :

https://www.beatthegmat.com/f-x-f-1-x-fu ... 13580.html
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Junior | Next Rank: 30 Posts
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by knowledgeisatreasure » Fri Jun 26, 2009 3:45 pm
thank you so much!!
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