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The function [tex]\( f \)[/tex] is defined as follows:

[tex]\[
f(x) = \begin{cases}
x + 6 & \text{if } -5 \leq x \ \textless \ 1 \\
8 & \text{if } x = 1 \\
-x + 3 & \text{if } x \ \textgreater \ 1
\end{cases}
\][/tex]

(a) Find the domain of the function.

The domain of the function [tex]\( f \)[/tex] is [tex]\(\square\)[/tex] (Type your answer in interval notation.)


Sagot :

To determine the domain of the piecewise function [tex]\( f(x) \)[/tex], we need to consider where the function is defined for each piece.

We have three pieces to consider:

1. [tex]\( f(x) = x + 6 \)[/tex] for [tex]\( -5 \leq x < 1 \)[/tex]
2. [tex]\( f(x) = 8 \)[/tex] for [tex]\( x = 1 \)[/tex]
3. [tex]\( f(x) = -x + 3 \)[/tex] for [tex]\( x > 1 \)[/tex]

Let's analyze each piece one by one:

1. For the first piece [tex]\( f(x) = x + 6 \)[/tex], the function is defined for [tex]\( x \)[/tex] in the interval [tex]\([-5, 1)\)[/tex]. This means [tex]\( x \)[/tex] can take any value from -5 (inclusive) to 1 (exclusive).

2. For the second piece [tex]\( f(x) = 8 \)[/tex], the function is defined specifically at [tex]\( x = 1 \)[/tex]. Thus, the function is defined at this single point.

3. For the third piece [tex]\( f(x) = -x + 3 \)[/tex], the function is defined for [tex]\( x \)[/tex] in the interval [tex]\( (1, \infty) \)[/tex]. This means [tex]\( x \)[/tex] can take any value greater than 1.

Combining all the intervals, we get the following:

- From [tex]\([-5, 1)\)[/tex] for the first piece,
- Plus the point [tex]\(\{1\}\)[/tex],
- Plus the interval [tex]\((1, \infty)\)[/tex] for the third piece.

Combining these intervals, we get the domain of the function as all real numbers starting at [tex]\(-5\)[/tex] and extending to [tex]\(\infty\)[/tex]:

[tex]\[ [-5, \infty) \][/tex]

Thus, the domain of the function [tex]\( f \)[/tex] is [tex]\( \boxed{[-5, \infty)} \)[/tex].
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