Answer:
[tex]\begin{gathered} f(-3)=2 \\ f(-1)=-3 \\ f(2)=0 \\ f(6)=2 \end{gathered}[/tex]
Explanation:
Given the piecewise defined function;
[tex]f(x)=\mleft\{\begin{aligned}-x-1for\rightarrow x<-1_{} \\ -3\text{for} \\ \sqrt[]{x-2}\text{ for}\rightarrow x\ge2\end{aligned}\mright.\rightarrow-1\leq x<2[/tex]
a) f(-3)
we want to find the value of f(x) at x=-3.
-3 is less than -1, so it falls within the interval x<-1.
[tex]\begin{gathered} f(x)=-x-1 \\ f(-3)=-(-3)-1 \\ f(-3)=3-1 \\ f(-3)=2 \end{gathered}[/tex]
b) f(-1)
-1 falls with the second interval
[tex]-1\leq x\leq2[/tex]
For this interval, f(x) is always equal to -3.
[tex]\begin{gathered} f(x)=-3 \\ f(-1)=-3 \end{gathered}[/tex]
c) f(2)
2 falls within the last interval.
[tex]\begin{gathered} f(x)=\sqrt[]{x-2} \\ f(2)=\sqrt[]{2-2} \\ f(2)=0 \end{gathered}[/tex]
d) f(6)
6 falls within the last interval.
[tex]\begin{gathered} f(x)=\sqrt[]{x-2} \\ f(6)=\sqrt[]{6-2} \\ f(6)=\sqrt[]{4} \\ f(6)=2 \end{gathered}[/tex]
