Let's fill in the values for [tex]\( q \)[/tex], [tex]\( r \)[/tex], [tex]\( s \)[/tex], and [tex]\( t \)[/tex] in the table based on the given information.
1. For [tex]\( x = -8 \)[/tex]:
- We need to use [tex]\( d(x) \)[/tex] because [tex]\( -8 \leq 0 \)[/tex].
- Given that [tex]\( d(x) = -\sqrt{\frac{1}{2} x + 4} \)[/tex]:
- The value of [tex]\( q \)[/tex] is [tex]\( d(-8) \)[/tex].
2. For [tex]\( x = 0 \)[/tex]:
- We need to use [tex]\( f(x) \)[/tex] because [tex]\( 0 \geq 0 \)[/tex].
- Given that [tex]\( f(x) = \sqrt{\frac{1}{2} x + 4} \)[/tex]:
- The value of [tex]\( r \)[/tex] is [tex]\( f(0) \)[/tex].
3. For [tex]\( x = 10 \)[/tex]:
- We need to use both [tex]\( f(x) \)[/tex] and [tex]\( d(x) \)[/tex]:
- The value of [tex]\( s \)[/tex] is [tex]\( f(10) \)[/tex].
- The value of [tex]\( t \)[/tex] is [tex]\( d(10) \)[/tex].
### Step-by-step solution:
For [tex]\( q \)[/tex] when [tex]\( x = -8 \)[/tex]:
- We use [tex]\( d(x) \)[/tex].
- Substituting [tex]\( x = -8 \)[/tex]:
[tex]\[ d(-8) = -\sqrt{\frac{1}{2}(-8) + 4} = -\sqrt{-4 + 4} = -\sqrt{0} = 0 \][/tex]
So, [tex]\( q = -0.0 \)[/tex] (or 0).
For [tex]\( r \)[/tex] when [tex]\( x = 0 \)[/tex]:
- We use [tex]\( f(x) \)[/tex].
- Substituting [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \sqrt{\frac{1}{2}(0) + 4} = \sqrt{0 + 4} = \sqrt{4} = 2 \][/tex]
So, [tex]\( r = 2.0 \)[/tex].
For [tex]\( s \)[/tex] when [tex]\( x = 10 \)[/tex]:
- We use [tex]\( f(x) \)[/tex].
- Substituting [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = \sqrt{\frac{1}{2}(10) + 4} = \sqrt{5 + 4} = \sqrt{9} = 3 \][/tex]
So, [tex]\( s = 3.0 \)[/tex].
For [tex]\( t \)[/tex] when [tex]\( x = 10 \)[/tex]:
- We use [tex]\( d(x) \)[/tex].
- Substituting [tex]\( x = 10 \)[/tex]:
[tex]\[ d(10) = -\sqrt{\frac{1}{2}(10) + 4} = -\sqrt{5 + 4} = -\sqrt{9} = -3 \][/tex]
So, [tex]\( t = -3.0 \)[/tex].
Based on the above calculations:
[tex]\[
\begin{array}{l}
q = -0.0 \\
r = 2.0 \\
s = 3.0 \\
t = -3.0 \\
\end{array}
\][/tex]
The completed table is:
[tex]\[
\begin{array}{|c|c|c|}
\hline x & f(x) & d(x) \\
\hline -8 & 0 & -0.0 \\
\hline 0 & 2.0 & -2 \\
\hline 10 & 3.0 & -3.0 \\
\hline
\end{array}
\][/tex]
