Functions, with inequality
It is given that f ( x ) is a function defined on R , satisfying f ( 1 ) = 1 , and for any x ∈ R ,
f ( x + 5 ) ≥ f ( x ) + 5
f ( x + 1 ) ≤ f ( x ) + 1
If g ( x ) = f ( x ) + 1 − x , what is the value of g ( 2 0 1 4 ) ?
The answer is 1.
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3 solutions
I put the logic in this way without constructing a function...
By induction we can prove f ( x ) ≤ x . Hence f ( 2 0 1 4 ) ≤ 2 0 1 4 .
and also f ( 2 0 0 9 ) ≥ 2 0 0 9
But we have f ( x + 5 ) ≥ x + 5 f ( 2 0 1 4 ) ≥ f ( 2 0 0 9 ) + 5 Putting the maximum value of f ( 2 0 0 9 ) we get f ( 2 0 1 4 ) ≥ 2 0 0 9 + 5 f ( 2 0 1 4 ) ≥ 2 0 1 4
Hence the only possibility is f ( 2 0 1 4 ) = 2 0 1 4
From the given: f(x + 5) >= f(x) + 5 implies f(6) >= 6. f(2) <= 2. Continuing the algorithm for the second inequality, f(3) <= f(2) + 1 <= 3, f(4) <= f(3) + 1 <= f(2) + 2 <= 4, f(5) <= f(4) + 1 <= f(3) + 2 <= f(2) + 3 <= 5, and f(6) <= f(5) + 1 <= f(4) + 2 <= f(3) + 3 <= f(2) + 4 <= 6. But f(6) >= 6, and so the only function that satisfies these inequalities is f(x) = x. Therefore, g(x) = f(x) - x + 1 = x - x + 1 = 1.
Using the definition of the function f(x)=x.We get the answer just by putting the value of x
g(x)=f(x)+1-x=f(2014)+1-2014=2014+1-2014=1
Note that applying rule 2 five times yields f ( x + 5 ) ≤ f ( x ) + 5 .
However, since f ( x + 5 ) ≥ f ( x ) + 5 by rule 1 , we must have f ( x + 5 ) = f ( x ) + 5 .
Now, we can see that we can let f ( x ) from x ∈ ( 1 , 6 ) be anything we want, as long as we make f ( x + 5 ) = f ( x ) + 5 . Thus, there is no definite answer .
However, assuming that f ( x ) = x because that is one of the infinitely many functions that satisfy the requirements, then g ( x ) = 1 for all x so g ( 2 0 1 4 ) = 1