If f(x) = \ array ll 2 + 2x, - 1 ≤ x < 0 1 - (x)/(3), 0 ≤ x ≤ 3...

Given: 

f\left(x\right) = \left\{$\begin{array}{ll}2 + 2x, x \in [ - 1, 0 ) \\ 1 - \dfrac{x}{3}, x \in \left[0, 3\right]\end{array}$\right) and g\left(x\right) = \left\{$\begin{array}{ll}x, x \in ( 0, 1 ] \\ - x, x \in \left[- 3, 0\right]\end{array}$\right)</p><p>⇒ f\left(g\left(x\right)\right) = \left\{$$\begin{array}{ll}2 + 2g\left(x\right), g\left(x\right) \in [ - 1, 0 ) \\ 1 - \dfrac{g\left(x\right)}{3}, g\left(x\right) \in \left[0, 3\right]\end{array}$$\right)

⇒ f\left(g\left(x\right)\right) = \left\{$\begin{array}{ll}2 + 2\left|x\right|, \left|x\right| \in [ - 1, 0 ) ⇒ x \in ϕ \\ 1 - \dfrac{\left|x\right|}{3}, \left|x\right| \in \left[0, 3\right]\end{array}$\right)</p><p>\left\{\text{as}x \in \left[- 3, 3\right]\text{for}\left|x\right| \le 3\text{but}x \in \left[- 3, 1\right]\text{in}g\left(x\right)\right\}$</p><p>$⇒ f\left(g\left(x\right)\right) = 1 - \dfrac{\left|x\right|}{3}, x \in \left[- 3, 1\right]$</p><p>Now, maximum value of$f\left(g\left(x\right)\right) = 1 - \dfrac{\left|x\right|}{3} = 1 - \dfrac{0}{3} = 1$</p><p>And minimum value is given by$f\left(g\left(x\right)\right) = 1 - \dfrac{3}{3} = 1 - 1 = 0$</p><p>Hence, the range of the function will be \left[0, 1\right]$