Prove that $\left | f(x) - a \right | \leq \frac{1}{2} \left | x - a \right |$

Question

$f(x) = 1 - \frac{1}{x}(\sqrt{1 + x^2}-1)$

$|f'(x)| \leq \frac{1}{2}$

$a$ is a solution for $f(x) = x$ where $0.65 < a < 0,7$

The question says:

Prove that $\left | f(x) - a \right | \leq \frac{1}{2} \left | x - a \right |$

I have no idea from where to begin, if I knew I would have posted my attempt at a solution but I didn't even know how to start it.

Any ideas?

Answer 1

Since $x = a$ is a solution to $f(x) = x$ , thus $f(a) = a$ . Now, take any $x \in \mathbb{R}$ . Without loss of generality, let $x \leqslant a$ . Then, by Lagrange's Mean Value Theorem, there exists $h_x \in (x , a)$ such that $$\left\vert \frac{f(x) - a}{x - a}\right\vert = \left\vert \frac{f(x) - f(a)}{x - a}\right\vert = \vert f'(h_x) \vert \leqslant \frac{1}{2}$$ which is what you require to show.

Similarly, if $x \geqslant a$ , then an exactly similar calculation will show the same result. Therefore, for any given $x \in \mathbb{R}$ , we'll have : $$\left | f(x) - a \right | \leq \frac{1}{2} \left | x - a \right |$$

Sorry if I didn't understand your question properly.