f:[-2,2] rightarrow[-2,2] and g:[-2,2] righta | vedclass

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(D) Given $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \ x^2-2, & 0 \leq x \leq 2 \end{cases}$.For $f(x)$, the range is $[-2, 2]$. Since $f(x) = -2$

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$f$ is not bijective mapping and $g$ is surjective mapping

Vedclass · Answer

(D) Given $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \ x^2-2, & 0 \leq x \leq 2 \end{cases}$.For $f(x)$, the range is $[-2, 2]$. Since $f(x) = -2$ for all $x \in [-2, 0]$, $f$ is not injective. Thus, $f$ is not bijective.Now, $g(x) = |f(x)| + f(|x|)$.If $-2 \leq x \leq 0$, then $|x| \in [0, 2]$, so $f(|x|) = |x|^2 - 2 = x^2 - 2$. Also $f(x) = -2$, so $|f(x)| = 2$. Thus $g(x) = 2 + x^2 - 2 = x^2$.If $0 \leq x \leq 2$, then $|x| = x$, so $f(|x|) = f(x) = x^2 - 2$. Thus $g(x) = |x^2 - 2| + x^2 - 2$.For $0 \leq x \leq \sqrt{2}$, $x^2 - 2 \leq 0$, so $g(x) = -(x^2 - 2) + x^2 - 2 = 0$.For $\sqrt{2}  0$, so $g(x) = (x^2 - 2) + x^2 - 2 = 2(x^2 - 2)$.Thus, $g(x) = \begin{cases} x^2, & -2 \leq x \leq 0 \ 0, & 0 \leq x \leq \sqrt{2} \ 2(x^2-2), & \sqrt{2} The range of $g(x)$ is $[0, 4]$, which is equal to the codomain, so $g$ is surjective. Since $g(x) = 0$ for $x \in [0, \sqrt{2}]$, $g$ is not injective.

$f:[-2,2] \rightarrow[-2,2]$ and $g:[-2,2] \rightarrow[0,4]$ are two functions defined as $f(x)=\begin{cases} -2, & -2 \leq x \leq 0 \\ x^2-2, & 0 \leq x \leq 2 \end{cases}$ and $g(x)=|f(x)|+f(|x|)$, then

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