1
$\begingroup$

How to prove that the function:

$g(x) = \frac{1}{f(x)}$

is bounded, when $f(x) \geq |x-1|$ and $f(1)=2$ and f(x) is continuous at the point $x=1$

3 Answers 3

1

You should split the set $\mathbb R$ into two areas:

The first area is the set $(1-\epsilon, 1+\epsilon)$. On that area, if you pick $\epsilon>0$ small enough, you can guarantee (from continuity of $f$), that $f$ will be greater than $1$.

On the other part, $(-\infty, 1-\epsilon]\cup[1+\epsilon, \infty)$, you can guarantee that $f(x)\geq |x-1|\geq \epsilon$, which means that $$|g(x)|=\left|\frac{1}{f(x)}\right| = \frac{1}{|f(x)|}\leq \frac1\epsilon.$$

(the last inequality following from the fact that if $a\geq b>0$, then $\frac1a\leq\frac1b$.

So, all together, if you find an appropriate $\epsilon$, you can show that $g$ is bounded by $\frac1\epsilon$

  • 0
    Do I understand the first part correctly if I say that any bounded epsilon >0 would satisfy the condition? – 2017-02-22
  • 0
    @Dole Certainly not. $f$ may actually be equal to $\frac12$ at $x=\frac12$, and in that case, $\epsilon=1$ would *not* be OK. – 2017-02-22
1

Suppose, to the contrary, that $g$ is not bounded. Then, to each $n \in \mathbb N$, there is $x_n$ such that $|g(x_n)| >n$. It follows:

(*) $|x_n-1| \le |f(x_n)| \le \frac{1}{n}$ for all $n$.

Hence $x_n \to 1$. Since $f$ is continuous at the point $x=1$, we get

$f(x_n) \to f(1)=2$.

But from (*) we derive $f(x_n) \to 0$, a contradiction.

0

Not sure , (since am a beginner and can't comment) but if g(x) = 1 / f(x) This implies g(x) < f(x) (if f(x) != 1) , this implies g(x) < |x-1| , and this pretty much bounds it!

  • 0
    eh sorry , x can be 2 , then f(x) can be 1 this reduces my answer to crap! – 2017-02-22