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

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(D) The function is defined as $f(x) = \begin{cases} 2x-3, & x  2 \end{cases}$.$1$. Checking for Injec

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neither injection nor surjection

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(D) The function is defined as $f(x) = \begin{cases} 2x-3, & x  2 \end{cases}$.$1$. Checking for Injectivity (One-to-One):$A$ function is injective if $f(x_1) = f(x_2) \implies x_1 = x_2$. Consider the interval $[-2, 2]$, where $f(x) = x^2 - 1$. We observe that $f(-1) = (-1)^2 - 1 = 0$ and $f(1) = (1)^2 - 1 = 0$. Since $f(-1) = f(1)$ but $-1 
eq 1$, the function is not injective.$2$. Checking for Surjectivity (Onto):$A$ function is surjective if its range equals its codomain $(R)$.- For $x - For $-2 \leq x \leq 2$, $f(x) = x^2 - 1$. The minimum value is $-1$ (at $x=0$) and the maximum is $f(-2) = f(2) = 3$. So, $f(x) \in [-1, 3]$.- For $x > 2$, $f(x) > 3(2) + 2 = 8$. So, $f(x) \in (8, \infty)$.The range of $f$ is $(-\infty, -7) \cup [-1, 3] \cup (8, \infty)$.Since the range is not equal to the codomain $R$, the function is not surjective.Conclusion: The function is neither injective nor surjective.

If the function $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} 2x-3, & \text{if } x < -2 \\ x^2-1, & \text{if } -2 \leq x \leq 2 \\ 3x+2, & \text{if } x > 2 \end{cases}$ then $f$ is

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