Answer:
[tex]f(-2)=4[/tex]
[tex]f(3) = \sf DNE[/tex]
[tex]f(0)=-3[/tex]
[tex]f(-5)=-2[/tex]
[tex]f(6)= 3[/tex]
Step-by-step explanation:
Given piecewise function:
[tex]f(x)=\begin{cases}2x+8\quad &\textsf{if}\; x\leq -2\\x^2-3&\textsf{if}\;-2 < x < 3\\\sqrt{x+3}&\textsf{if}\;x > 3\end{cases}[/tex]
To find the values of f(x), we need to use the definitions of the function f(x) in the given intervals.
For f(-2), as x = -2 falls within the interval x ≤ -2, we need to use f(x) = 2x + 8:
[tex]f(-2)=2(-2)+8\\\\f(-2)=-4+8\\\\f(-2)=4[/tex]
As x = 3 is not included in any of the specified intervals, the value f(3) does not exist. So:
[tex]f(3) = \sf DNE[/tex]
For f(0), as x = 0 falls within the interval -2 < x < 3, we need to use f(x) = x² - 3:
[tex]f(0)=0^2-3\\\\f(0)=0-3\\\\f(0)=-3[/tex]
For f(-5), as x = -5 falls within the interval x ≤ -2, we need to use f(x) = 2x + 8:
[tex]f(-5)=2(-5)+8\\\\f(-5)=-10+8\\\\f(-5)=-2[/tex]
For f(6), as x = 6 falls within the interval x > 3, we need to use f(x) = √(x + 3):
[tex]f(6)=\sqrt{6+3}\\\\f(6)=\sqrt{9}\\\\f(6)=3[/tex]
Additional Notes
The graph of the piecewise function is attached, which can be used to verify all the values in the question. Note that an open circle signifies that the value of x is not included in the interval, while a closed circle indicates that the value of x is included in the interval.
