Define f: R rightarrow R by f(x) = max {x+1, | vedclass

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(C) Given $f(x) = \max \{x+1, 1-x, 2\}$.We can analyze the function by breaking it into intervals:For $x  2$,so $f(x) = 1-x$

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neither one-one nor onto

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(C) Given $f(x) = \max \{x+1, 1-x, 2\}$.We can analyze the function by breaking it into intervals:For $x  2$,so $f(x) = 1-x$.For $-1 \leq x \leq 1$,$x+1 \geq 0$,$1-x \geq 0$,and the maximum of these and $2$ is $2$ (since $x+1 \leq 2$ and $1-x \leq 2$ for $x \in [-1, 1]$).For $x > 1$,$x+1 > 2$ and $1-x Thus,$f(x) = \begin{cases} 1-x, & x  1 \end{cases}$.Since $f(x) = 2$ for all $x \in [-1, 1]$,the function is not one-one (many-to-one).Since the range of $f(x)$ is $[2, \infty)$,which is a proper subset of the codomain $R$,the function is not onto.Therefore,$f$ is neither one-one nor onto.

Define $f: R \rightarrow R$ by $f(x) = \max \{x+1, 1-x, 2\}$. Then,$f$ is

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