Option B:
f(3) = 5
Solution:
Given function:
[tex]f(x)=\left\{\begin{aligned}-x^{2}, \ \ & x<-2 \\3, \ \ &-2 \leq x<0 \\x+2, \ \ & x \geq 0\end{aligned}\right.[/tex]
- If x value is less than –2, that is –3, –4, –5, –6, ... then f(x) = –x².
- If x value is greater than or equal to –2 and less than 0, that is –2 and –1 then f(x) = 3.
- If x value is greater than or equal to 0, that is 0, 1, 2, 3,... then f(x) = x +2.
To find the value of f(3):
Here x value is 3 which is greater than 0 (3 > 0).
Therefore, f(x) = x + 2.
Substitute x = 3 in the above function.
f(3) = 3 + 2
f(3) = 5
Hence option B is the correct answer.
