If the function f: R rightarrow R is defined | vedclass

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(C) Given,$f(x)=x|x|$.We can write this as:$f(x) = \begin{cases} x^2, & x \geq 0 \ -x^2, & x < 0 \end{cases}$$1$. One-one check:Let $f(x_1) = f(x_2)$

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$f$ is both one-one and onto

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(C) Given,$f(x)=x|x|$.We can write this as:$f(x) = \begin{cases} x^2, & x \geq 0 \ -x^2, & x $1$. One-one check:Let $f(x_1) = f(x_2)$.If $x_1, x_2 \geq 0$,then $x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (since $x \geq 0$).If $x_1, x_2 If one is positive and one is negative,$f(x)$ will have different signs,so $f(x_1) 
eq f(x_2)$.Thus,$f(x)$ is one-one.$2$. Onto check:For any $y \in R$,we can find $x$ such that $f(x) = y$.If $y \geq 0$,$x = \sqrt{y}$. If $y Since for every $y$ in the codomain $R$,there exists an $x$ in the domain $R$,the function is onto.Therefore,$f$ is both one-one and onto.

If the function $f: R \rightarrow R$ is defined by $f(x)=x|x|$,then:

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