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In Exercises 37-42, evaluate each piecewise function at the given values of the independent variable.

37. [tex]f(x)=\left\{\begin{array}{lll}3x + 5 & \text{if} & x \ \textless \ 0 \\ 4x + 7 & \text{if} & x \geq 0\end{array}\right.[/tex]

a. [tex]f(-2)[/tex]
b. [tex]f(0)[/tex]
c. [tex]f(3)[/tex]

Answer :

Sure, let's evaluate each piecewise function at the given values of the independent variable, following each step methodically.

The piecewise function given is:
[tex]\[ f(x) = \begin{cases}
3x + 5 & \text{if } x < 0 \\
4x + 7 & \text{if } x \geq 0
\end{cases} \][/tex]

Now we will evaluate this function at [tex]\(x = -2\)[/tex], [tex]\(x = 0\)[/tex], and [tex]\(x = 3\)[/tex].

### a. Evaluate [tex]\(f(-2)\)[/tex]
For [tex]\(x = -2\)[/tex]:
- Since [tex]\(-2 < 0\)[/tex], we use the first part of the piecewise function, [tex]\(3x + 5\)[/tex].

[tex]\[ f(-2) = 3(-2) + 5 \][/tex]
[tex]\[ f(-2) = -6 + 5 \][/tex]
[tex]\[ f(-2) = -1 \][/tex]

Therefore, [tex]\( f(-2) = -1 \)[/tex].

### b. Evaluate [tex]\(f(0)\)[/tex]
For [tex]\(x = 0\)[/tex]:
- Since [tex]\(0 \ge 0\)[/tex], we use the second part of the piecewise function, [tex]\(4x + 7\)[/tex].

[tex]\[ f(0) = 4(0) + 7 \][/tex]
[tex]\[ f(0) = 0 + 7 \][/tex]
[tex]\[ f(0) = 7 \][/tex]

Therefore, [tex]\( f(0) = 7 \)[/tex].

### c. Evaluate [tex]\(f(3)\)[/tex]
For [tex]\(x = 3\)[/tex]:
- Since [tex]\(3 \ge 0\)[/tex], we use the second part of the piecewise function, [tex]\(4x + 7\)[/tex].

[tex]\[ f(3) = 4(3) + 7 \][/tex]
[tex]\[ f(3) = 12 + 7 \][/tex]
[tex]\[ f(3) = 19 \][/tex]

Therefore, [tex]\( f(3) = 19 \)[/tex].

So the values are:
- [tex]\( f(-2) = -1 \)[/tex]
- [tex]\( f(0) = 7 \)[/tex]
- [tex]\( f(3) = 19 \)[/tex]

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