Let A = {x in R : -1 leq x leq 1} and f: A ri | vedclass

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(D) The function is defined as $f(x) = x|x|$.We can write this as a piecewise function:$f(x) = \begin{cases} -x^2, & -1 \leq x < 0 \ x^2, & 0 \leq x

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bijective

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(D) The function is defined as $f(x) = x|x|$.We can write this as a piecewise function:$f(x) = \begin{cases} -x^2, & -1 \leq x For injectivity: Since $f(x)$ is a strictly increasing function on the interval $[-1, 1]$ (as its derivative $f'(x) = 2|x| \geq 0$), it is injective.For surjectivity: The range of $f(x)$ for $x \in [-1, 0)$ is $(-1, 0]$ and for $x \in [0, 1]$ is $[0, 1]$. Combining these, the range is $[-1, 1]$, which is equal to the codomain $A$. Thus, the function is surjective.Since the function is both injective and surjective, it is bijective.

Let $A = \{x \in R : -1 \leq x \leq 1\}$ and $f: A \rightarrow A$ be a mapping defined by $f(x) = x|x|$. Then $f$ is

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