Define the function [tex]f(x)[/tex] as follows:

[tex]\[f(x) = \begin{cases}
-x^2 + 4 & \text{if } -3 \leq x \ \textless \ 0 \\
4 & \text{if } 0 \leq x \ \textless \ 3 \\
-x + 5 & \text{if } x \geq 3
\end{cases}\][/tex]



Answer :

Sure, let's evaluate the function [tex]\( f(x) \)[/tex] for specific values of [tex]\( x \)[/tex]. We have three cases depending on the value of [tex]\( x \)[/tex].

1. For [tex]\( -3 \leq x < 0 \)[/tex], we have [tex]\( f(x) = -x^2 + 4 \)[/tex].
2. For [tex]\( 0 \leq x < 3 \)[/tex], we have [tex]\( f(x) = 4 \)[/tex].
3. For [tex]\( x \geq 3 \)[/tex], we have [tex]\( f(x) = -x + 5 \)[/tex].

Now, let's evaluate [tex]\( f(x) \)[/tex] at specific values of [tex]\( x \)[/tex]:

1. Evaluate at [tex]\( x = -3 \)[/tex]:
- Since [tex]\( -3 \leq -3 < 0 \)[/tex], we use [tex]\( f(x) = -x^2 + 4 \)[/tex].
[tex]\[ f(-3) = -(-3)^2 + 4 = -9 + 4 = -5 \][/tex]

2. Evaluate at [tex]\( x = -1 \)[/tex]:
- Since [tex]\( -3 \leq -1 < 0 \)[/tex], we use [tex]\( f(x) = -x^2 + 4 \)[/tex].
[tex]\[ f(-1) = -(-1)^2 + 4 = -1 + 4 = 3 \][/tex]

3. Evaluate at [tex]\( x = 0 \)[/tex]:
- Since [tex]\( 0 \leq 0 < 3 \)[/tex], we use [tex]\( f(x) = 4 \)[/tex].
[tex]\[ f(0) = 4 \][/tex]

4. Evaluate at [tex]\( x = 2 \)[/tex]:
- Since [tex]\( 0 \leq 2 < 3 \)[/tex], we use [tex]\( f(x) = 4 \)[/tex].
[tex]\[ f(2) = 4 \][/tex]

5. Evaluate at [tex]\( x = 3 \)[/tex]:
- Since [tex]\( x \geq 3 \)[/tex], we use [tex]\( f(x) = -x + 5 \)[/tex].
[tex]\[ f(3) = -3 + 5 = 2 \][/tex]

6. Evaluate at [tex]\( x = 4 \)[/tex]:
- Since [tex]\( x \geq 3 \)[/tex], we use [tex]\( f(x) = -x + 5 \)[/tex].
[tex]\[ f(4) = -4 + 5 = 1 \][/tex]

So, the evaluations result in the following:

- [tex]\( f(-3) = -5 \)[/tex]
- [tex]\( f(-1) = 3 \)[/tex]
- [tex]\( f(0) = 4 \)[/tex]
- [tex]\( f(2) = 4 \)[/tex]
- [tex]\( f(3) = 2 \)[/tex]
- [tex]\( f(4) = 1 \)[/tex]

Thus, the calculated outputs for the function [tex]\( f(x) \)[/tex] at the specified values are:

[tex]\[ [-5, 3, 4, 4, 2, 1] \][/tex]

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