Let's determine the values of [tex]\(q\)[/tex], [tex]\(r\)[/tex], [tex]\(s\)[/tex], and [tex]\(t\)[/tex] using the given inverse functions [tex]\(f(x)\)[/tex] and [tex]\(d(x)\)[/tex].
Given:
- For [tex]\(x \geq 0\)[/tex], the inverse function is [tex]\(f(x)=\sqrt{\frac{1}{2} x + 4}\)[/tex]
- For [tex]\(x \leq 0\)[/tex], the inverse function is [tex]\(d(x)=-\sqrt{\frac{1}{2} x + 4}\)[/tex]
We need to calculate the following:
1. [tex]\( q = d(-8) \)[/tex]
2. [tex]\( r = f(0) \)[/tex]
3. [tex]\( s = f(10) \)[/tex]
4. [tex]\( t = d(10) \)[/tex]
### Step-by-Step Calculation:
#### 1. [tex]\( q = d(-8) \)[/tex]
For [tex]\(x = -8\)[/tex], use the function [tex]\(d(x)\)[/tex]:
[tex]\[ d(x) = -\sqrt{\frac{1}{2} x + 4} \][/tex]
Substitute [tex]\(x = -8\)[/tex]:
[tex]\[ d(-8) = -\sqrt{\frac{1}{2} (-8) + 4} \][/tex]
[tex]\[ = -\sqrt{\frac{-8}{2} + 4} \][/tex]
[tex]\[ = -\sqrt{-4 + 4} \][/tex]
[tex]\[ = -\sqrt{0} \][/tex]
[tex]\[ = -0 \][/tex]
Thus, [tex]\( q = -0 \)[/tex].
#### 2. [tex]\( r = f(0) \)[/tex]
For [tex]\(x = 0\)[/tex], use the function [tex]\(f(x)\)[/tex]:
[tex]\[ f(x) = \sqrt{\frac{1}{2} x + 4} \][/tex]
Substitute [tex]\(x = 0\)[/tex]:
[tex]\[ f(0) = \sqrt{\frac{1}{2} (0) + 4} \][/tex]
[tex]\[ = \sqrt{0 + 4} \][/tex]
[tex]\[ = \sqrt{4} \][/tex]
[tex]\[ = 2 \][/tex]
Thus, [tex]\( r = 2 \)[/tex].
#### 3. [tex]\( s = f(10) \)[/tex]
For [tex]\(x = 10\)[/tex], use the function [tex]\(f(x)\)[/tex]:
[tex]\[ f(x) = \sqrt{\frac{1}{2} x + 4} \][/tex]
Substitute [tex]\(x = 10\)[/tex]:
[tex]\[ f(10) = \sqrt{\frac{1}{2} (10) + 4} \][/tex]
[tex]\[ = \sqrt{\frac{10}{2} + 4} \][/tex]
[tex]\[ = \sqrt{5 + 4} \][/tex]
[tex]\[ = \sqrt{9} \][/tex]
[tex]\[ = 3 \][/tex]
Thus, [tex]\( s = 3 \)[/tex].
#### 4. [tex]\( t = d(10) \)[/tex]
For [tex]\(x = 10\)[/tex], use the function [tex]\(d(x)\)[/tex]:
[tex]\[ d(x) = -\sqrt{\frac{1}{2} x + 4} \][/tex]
Substitute [tex]\(x = 10\)[/tex]:
[tex]\[ d(10) = -\sqrt{\frac{1}{2} (10) + 4} \][/tex]
[tex]\[ = -\sqrt{\frac{10}{2} + 4} \][/tex]
[tex]\[ = -\sqrt{5 + 4} \][/tex]
[tex]\[ = -\sqrt{9} \][/tex]
[tex]\[ = -3 \][/tex]
Thus, [tex]\( t = -3 \)[/tex].
### Final Values:
[tex]\[
\begin{array}{l}
q = -0 \\
r = 2 \\
s = 3 \\
t = -3 \\
\end{array}
\][/tex]
So the completed table is:
[tex]\[
\begin{tabular}{|c|c|c|}
\hline$x$ & $f(x)$ & $d(x)$ \\
\hline-8 & 0 & -0 \\
\hline 0 & 2 & -2 \\
\hline 10 & 3 & -3 \\
\hline
\end{tabular}
\][/tex]
