Let f : R rightarrow R be a function such tha | vedclass

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(C) Given $f(x) = \frac{x^2+2x+1}{x^2+1} = \frac{(x^2+1) + 2x}{x^2+1} = 1 + \frac{2x}{x^2+1}$.To check for one-one or many-one,we find the derivative

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$f(x)$ is one-one in $[1, \infty)$ but not in $(-\infty, \infty)$

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(C) Given $f(x) = \frac{x^2+2x+1}{x^2+1} = \frac{(x^2+1) + 2x}{x^2+1} = 1 + \frac{2x}{x^2+1}$.To check for one-one or many-one,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx} \left( 1 + \frac{2x}{x^2+1} \right) = \frac{(x^2+1)(2) - (2x)(2x)}{(x^2+1)^2} = \frac{2x^2+2-4x^2}{(x^2+1)^2} = \frac{2-2x^2}{(x^2+1)^2} = \frac{2(1-x)(1+x)}{(x^2+1)^2}$.Critical points are $x = 1$ and $x = -1$.For $x \in (1, \infty)$,$f'(x) For $x \in (-1, 1)$,$f'(x) > 0$,so the function is strictly increasing.For $x \in (-\infty, -1)$,$f'(x) Since the function is strictly decreasing in $(-\infty, -1)$ and $(1, \infty)$,and strictly increasing in $(-1, 1)$,it is one-one in these intervals. However,it is not one-one in $(-\infty, \infty)$ because $f(x)$ takes the same values at different points (e.g.,$f(0) = 1$ and $f(\infty) = 1$).Thus,$f(x)$ is one-one in $[1, \infty)$ but not in $(-\infty, \infty)$.

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

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