normacruz1554
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Function: [tex]\( g(x) = 2x^2 - 8 \)[/tex]

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

For [tex]\( x \leq 0 \)[/tex], the inverse function is [tex]\( d(x) = -\sqrt{\frac{x}{2} + 4} \)[/tex]

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

Sagot :

GinnyAnswer
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]
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