Math Is Fun Forum

harrychess

Member

Registered: 2014-04-04

Posts: 33

Confusing Function Problem

Let f be a function such that f(x+y) = x + f(y) for any two real numbers x and y. If f(0) = 2, then what is f(2012)? c

How are you supposed to solve this? Plus, I am really confused on how you are supposed to start this problem.

zetafunc.

Guest

Re: Confusing Function Problem

Can you think of any helpful substitutions?

harrychess

Member

Registered: 2014-04-04

Posts: 33

Re: Confusing Function Problem

Never mind, I found the answer. 2014

harrychess

Member

Registered: 2014-04-04

Posts: 33

Re: Confusing Function Problem

f(2012)=f(2012+0)

Then this becomes 2012+f(0). We know that f(0) is 2 so f(2012) is 2014.

harrychess

Member

Registered: 2014-04-04

Posts: 33

Re: Confusing Function Problem

Here's another question.

Let p(x) = 2x^3  - 113 and let q be the inverse of p. Find q(137).

Aren't you supposed to set 2x^3  - 113=137 or something like that?

harrychess

Member

Registered: 2014-04-04

Posts: 33

Re: Confusing Function Problem

I just think I figured it out. Is it 5?

The inverse of

is

Then, we just plug 137 for x into function q.

So,

equals 5.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Confusing Function Problem

Hi;

I have fixed your latex. here we do not use the $ tags but use the math tags instead.

Do you know why the two complex answers are rejected?

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
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