Let f(x) = begin{cases} x, & x

Question

(B) Given $f(x) = \begin{cases} x, & x < 0 \ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$.Since $x - [x] = \{x\}$,where $\{x\}$ is the f

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$[2, 5)$

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(B) Given $f(x) = \begin{cases} x, & x Since $x - [x] = \{x\}$,where $\{x\}$ is the fractional part function,we have $g(x) = 1 + \{x\}$.We know that for any $x \in \mathbb{R}$,$0 \leq \{x\} Therefore,$1 \leq 1 + \{x\} Since $g(x) \geq 1$ for all $x \in \mathbb{R}$,we use the definition of $f(x)$ for $x \geq 0$ to evaluate $f(g(x))$.$f(g(x)) = 1 + (g(x))^2 = 1 + (1 + \{x\})^2$.Since $0 \leq \{x\} Squaring the inequality,we get $1^2 \leq (1 + \{x\})^2 Adding $1$ to all parts,we get $1 + 1 \leq 1 + (1 + \{x\})^2 Thus,$2 \leq f(g(x)) The range of $f(g(x))$ is $[2, 5)$.

Let $f(x) = \begin{cases} x, & x < 0 \\ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$,then the range of $f(g(x))$ is (where $[.]$ denotes the greatest integer function).

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