Let $$f : R \to R$$ be a function such that $$f(x) = \frac{x^2+2x+1}{x^2+1}$$. Then

Given the function f(x) as

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

To find the behaviour of the f(x), we need to first calculate its derivative :

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

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

$$f'(x) = 0 \implies x = \pm 1$$ 

For $$x < -1, f'(x) < 0 \implies f(x)$$ is decreasing from $$1$$ to $$0$$

For $$-1 < x < 1, f'(x) > 0 \implies f(x)$$ is increasing from $$0$$ to $$2$$

For $$x > 1, f'(x) < 0 \implies f(x)$$ is decreasing from $$2$$ to $$1$$ 

But overall, we can see that $$f(x)$$ is many-one in $$R$$ because $$f'(x)$$ changes sign at $$x = -1$$ and $$x = 1$$ .

Specifically, for the interval $$[1, \infty)$$ , $$f'(x) \leq 0$$ (it is strictly decreasing for $$x > 1$$ ).

Thus, $$f(x)$$ is one-one in $$[1, \infty)$$ .