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by Night reader » Thu Jan 20, 2011 11:16 pm
yellowho wrote:The function f(x) is such that f(3x+1) = 3f(x) + 1 for all values of x. If f(2)=5, then f(7)=?
3f(x)=f(3x+1)-1 -> 3f(2)=f(6+1)-1 -> 3*5=f(7)-1 -> 16=f(7)
Last edited by Night reader on Fri Jan 21, 2011 3:35 am, edited 1 time in total.
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by yellowho » Thu Jan 20, 2011 11:26 pm
I don't know what you did there but I think you got the relationship backward.

My real question in this is, is there a way to solve this if 3x+1=7 is not true (X is not 2) as defined? Can you look at this like a sequence?
Last edited by yellowho on Thu Jan 20, 2011 11:38 pm, edited 1 time in total.
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by Everest » Thu Jan 20, 2011 11:36 pm
The function f(x) is such that f(3x+1) = 3f(x) + 1 for all values of x. If f(2)=5, then f(7)=?


f(3x+1) = 3f(x)+1 => f(7) = f(3*2+1)

= 3 f(2) + 1 = 3*5+1 = 16
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by Night reader » Thu Jan 20, 2011 11:49 pm
yellowho wrote:I don't know what you did there but I think you got the relationship backward.

My real question in this is, is there a way to solve this if 3x+1=7 is not true (X is not 2) as defined? Can you look at this like a sequence?
f(3x+1) = 3f(x) + 1 not 3x+1=7
f(2)=5 is different function
we need to find yet another function f(7)

I tried my best, though it's nonstandard algebra application rather than backward restatement.
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by Rahul@gurome » Thu Jan 20, 2011 11:52 pm
yellowho wrote:The function f(x) is such that f(3x+1) = 3f(x) + 1 for all values of x. If f(2)=5, then f(7)=?
We know f(2) and our target is to determine the value of f(7). How we can do that?

One quiet obvious way is to express f(7) in terms of f(2). Can we do that? Yes! From the relations given, we can do that easily. How?

Just put x = 2 in the given relation.
Now we have, f(3x + 1) = 3f(x) + 1
=> f(3*2 + 1) = 3f(2) + 1
=> f(7) = 3f(2) + 1

Now we have f(7) in terms of f(2).
Just put the value of f(2), and we have f(7) = 3*5 + 1 = 16
Night Reader wrote:f(2)=f(6+1)=3f(2)+1 = 5 -> f(7)=3f(2)+1-5 -> f(7)=3f(2)-4
How will you justify that step?
What you've wrote is equivalent to f(2) = f(7)
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by yellowho » Thu Jan 20, 2011 11:53 pm
sorry didnt mean to offend. i just didn't understand what u did. that doesnt mean you are wrong tho. i posted this question to look for different ways to solve so a different approach is welcome.
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by Night reader » Fri Jan 21, 2011 3:36 am
Rahul@gurome wrote:
yellowho wrote:The function f(x) is such that f(3x+1) = 3f(x) + 1 for all values of x. If f(2)=5, then f(7)=?
We know f(2) and our target is to determine the value of f(7). How we can do that?

One quiet obvious way is to express f(7) in terms of f(2). Can we do that? Yes! From the relations given, we can do that easily. How?

Just put x = 2 in the given relation.
Now we have, f(3x + 1) = 3f(x) + 1
=> f(3*2 + 1) = 3f(2) + 1
=> f(7) = 3f(2) + 1

Now we have f(7) in terms of f(2).
Just put the value of f(2), and we have f(7) = 3*5 + 1 = 16
Night Reader wrote:f(2)=f(6+1)=3f(2)+1 = 5 -> f(7)=3f(2)+1-5 -> f(7)=3f(2)-4
How will you justify that step?
What you've wrote is equivalent to f(2) = f(7)
thanks Rahul, I have edited my solution.
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