0
$\begingroup$

Suppose $f$ is a real-valued function and $f(x) \geq 1$ for every real $x \in [0,1]$.

Why we can always find a unique positive integer $n$ such that $2^{n} \leq f(x) < 2^{n+1}$ ?

  • 0
    @User: Take $f(x)=e^{x}$. Clearly $f(x) \geq 1$ for every $x \in [0,1]$. But as $x \to \infty$, $e^{x} \to \infty$.2010-11-17
  • 0
    Do you mean for any given $x$? Or for all $x$ at once? The second statement is false; the first is true because $f(x)\ge 1=2^0$ and $\{[2^n,2^{n+1})\}$ partitions $[1,\infty)$.2010-11-17

2 Answers 2

1

The intervals $[2^n,2^{n+1})$ cover the entire real half (existence) line $y\geq 1$ and no two of them intersect (uniqueness). So any point on the graph of the function must intersect one of these since $f(x)\geq 1$. You don't need any hypothesis on the domain of the function.

0

You can't do it for $f(x)=3x+1$.