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

For [tex]$x \geq 0$[/tex], the inverse function is [tex]$f(x)=\sqrt{\frac{1}{2}x+4}$[/tex]

For [tex][tex]$x \leq 0$[/tex][/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 :

To determine the values of [tex]\( q \)[/tex], [tex]\( r \)[/tex], [tex]\( s \)[/tex], and [tex]\( t \)[/tex], we need to correctly evaluate the inverse functions [tex]\( f(x) \)[/tex] and [tex]\( d(x) \)[/tex] at the given [tex]\(x\)[/tex] values as outlined in the problem statement.

1. Evaluate [tex]\( d(x) \)[/tex] at [tex]\( x = -8 \)[/tex] to find [tex]\( q \)[/tex]:

Since [tex]\( x = -8 \leq 0 \)[/tex], we use [tex]\( d(x) = -\sqrt{\frac{1}{2} x + 4} \)[/tex]:
[tex]\[ d(-8) = -\sqrt{\frac{1}{2} (-8) + 4} = -\sqrt{-4 + 4} = -\sqrt{0} = -0.0 \][/tex]
So, [tex]\( q = -0.0 \)[/tex].

2. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex] to find [tex]\( r \)[/tex]:

Since [tex]\( x = 0 \geq 0 \)[/tex], we use [tex]\( f(x) = \sqrt{\frac{1}{2} x + 4} \)[/tex]:
[tex]\[ f(0) = \sqrt{\frac{1}{2} (0) + 4} = \sqrt{0 + 4} = \sqrt{4} = 2.0 \][/tex]
So, [tex]\( r = 2.0 \)[/tex].

3. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 10 \)[/tex] to find [tex]\( s \)[/tex]:

Since [tex]\( x = 10 \geq 0 \)[/tex], we use [tex]\( f(x) = \sqrt{\frac{1}{2} x + 4} \)[/tex]:
[tex]\[ f(10) = \sqrt{\frac{1}{2} (10) + 4} = \sqrt{5 + 4} = \sqrt{9} = 3.0 \][/tex]
So, [tex]\( s = 3.0 \)[/tex].

4. Evaluate [tex]\( d(x) \)[/tex] at [tex]\( x = 10 \)[/tex] to find [tex]\( t \)[/tex]:

Although [tex]\( x = 10 \geq 0 \)[/tex], the problem explicitly specifies evaluating [tex]\( d(x) \)[/tex], therefore:
[tex]\[ d(10) = -\sqrt{\frac{1}{2} (10) + 4} = -\sqrt{5 + 4} = -\sqrt{9} = -3.0 \][/tex]
So, [tex]\( t = -3.0 \)[/tex].

In summary:

[tex]\[ \begin{array}{l} q = -0.0 \\ r = 2.0 \\ s = 3.0 \\ t = -3.0 \\ \end{array} \][/tex]

So the completed table with the calculated values 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]

And the values:

[tex]\[ q = -0.0 \quad r = 2.0 \quad s = 3.0 \quad t = -3.0 \][/tex]