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Function: [tex]g(x)=2 x^2-8[/tex]

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].

[tex]\[
\begin{tabular}{|c|c|c|}
\hline
x & f(x) & d(x) \\
\hline
-8 & 0 & q \\
\hline
0 & r & -2 \\
\hline
10 & s & t \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{array}{l}
q = \square \\
r = \square \\
s = \square \\
t = \square
\end{array}
\][/tex]



Answer :

GinnyAnswer
Let's solve for the values of \(q\), \(r\), \(s\), and \(t\) step-by-step using the given functions and the table.

1. Calculate \(q\):
[tex]\[ q = d(-8) \][/tex]
Given \(d(x) = -\sqrt{\frac{1}{2} x + 4}\),
[tex]\[ d(-8) = -\sqrt{\frac{1}{2} (-8) + 4} = -\sqrt{-4 + 4} = -\sqrt{0} = -0 \][/tex]
Thus,
[tex]\[ q = -0 \quad \text{(or just 0, but for consistency we'll use -0)} \][/tex]

2. Calculate \(r\):
[tex]\[ r = f(0) \][/tex]
Given \(f(x) = \sqrt{\frac{1}{2} x + 4}\),
[tex]\[ f(0) = \sqrt{\frac{1}{2} (0) + 4} = \sqrt{0 + 4} = \sqrt{4} = 2 \][/tex]
Thus,
[tex]\[ r = 2 \][/tex]

3. Calculate \(s\):
[tex]\[ s = f(10) \][/tex]
Given \(f(x) = \sqrt{\frac{1}{2} x + 4}\),
[tex]\[ f(10) = \sqrt{\frac{1}{2} (10) + 4} = \sqrt{5 + 4} = \sqrt{9} = 3 \][/tex]
Thus,
[tex]\[ s = 3 \][/tex]

4. Calculate \(t\):
[tex]\[ t = d(10) \][/tex]
Given \(d(x) = -\sqrt{\frac{1}{2} x + 4}\),
[tex]\[ d(10) = -\sqrt{\frac{1}{2} (10) + 4} = -\sqrt{5 + 4} = -\sqrt{9} = -3 \][/tex]
Thus,
[tex]\[ t = -3 \][/tex]

Collating these results, we get:
[tex]\[ \begin{array}{l} q = -0 \\ r = 2 \\ s = 3 \\ t = -3 \end{array} \][/tex]

So, the completed table is:

[tex]\[ \begin{array}{|c|c|c|} \hline x & f(x) & d(x) \\ \hline -8 & 0 & -0 \\ \hline 0 & 2 & -2 \\ \hline 10 & 3 & -3 \\ \hline \end{array} \][/tex]

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