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Evaluate the following piecewise function:

Given:
[tex]\[ f(x)=\left\{\begin{array}{ll}
7x + 3, & \text{if } x \ \textless \ 0 \\
7x + 6, & \text{if } x \ge 0
\end{array}\right. \][/tex]

1. [tex]\( f(-1) \)[/tex]
2. [tex]\( f(0) \)[/tex]
3. [tex]\( f(2) \)[/tex]

Sagot :

GinnyAnsweridn
Sure, let's evaluate the piecewise function step-by-step for the given values.

The given piecewise function is:
[tex]\[ f(x) = \begin{cases} 7x + 3, & \text{if } x < 0 \\ 7x + 6, & \text{if } x > 0 \end{cases} \][/tex]

1. Evaluating [tex]\( f(-1) \)[/tex]:
- Since [tex]\(-1\)[/tex] is less than [tex]\(0\)[/tex] ([tex]\(x < 0\)[/tex]), we use the first part of the piecewise function: [tex]\(7x + 3\)[/tex].
- Substituting [tex]\(x = -1\)[/tex]:
[tex]\[ f(-1) = 7(-1) + 3 = -7 + 3 = -4 \][/tex]

2. Evaluating [tex]\( f(0) \)[/tex]:
- For [tex]\(x = 0\)[/tex], the function is not defined in the given piecewise form as it only specifies conditions for [tex]\(x < 0\)[/tex] and [tex]\(x > 0\)[/tex], not for [tex]\(x = 0\)[/tex].
- Therefore, [tex]\( f(0) \)[/tex] is undefined:
[tex]\[ f(0) = \text{None} \][/tex]

3. Evaluating [tex]\( f(2) \)[/tex]:
- Since [tex]\(2\)[/tex] is greater than [tex]\(0\)[/tex] ([tex]\(x > 0\)[/tex]), we use the second part of the piecewise function: [tex]\(7x + 6\)[/tex].
- Substituting [tex]\(x = 2\)[/tex]:
[tex]\[ f(2) = 7(2) + 6 = 14 + 6 = 20 \][/tex]

So, the evaluated results are:
- [tex]\(f(-1) = -4\)[/tex]
- [tex]\(f(0) =\)[/tex] \text{None} (undefined)
- [tex]\(f(2) = 20\)[/tex]
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