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[tex][tex]$f(x)$[/tex][/tex] and [tex][tex]$g(x)$[/tex][/tex] are continuous functions. The table below lists the values of [tex][tex]$f(x)$[/tex][/tex], [tex][tex]$g(x)$[/tex][/tex], and their derivatives for certain values of [tex][tex]$x$[/tex][/tex].

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] & [tex]$g(x)$[/tex] & [tex]$f^{\prime}(x)$[/tex] & [tex]$g^{\prime}(x)$[/tex] \\
\hline
-5 & 8 & -6 & 9 & 1 \\
\hline
1 & -5 & 4 & 2 & 6 \\
\hline
2 & -5 & -5 & 4 & -6 \\
\hline
4 & 1 & -5 & -4 & 5 \\
\hline
6 & 8 & -1 & 1 & -9 \\
\hline
\end{tabular}
\][/tex]

Use the values in the table to find [tex][tex]$h^{\prime}(1)$[/tex][/tex] where [tex][tex]$h(x) = f(x) \times g(x)$[/tex][/tex].

Round the decimal answer to three decimal places if needed.



Answer :

GinnyAnswer
To find [tex]\( h'(1) \)[/tex] where [tex]\( h(x) = f(x) \cdot g(x) \)[/tex], we need to use the product rule for differentiation. The product rule states:

[tex]\[ h'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) \][/tex]

Given the values from the table for [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -5 \][/tex]
[tex]\[ g(1) = 4 \][/tex]
[tex]\[ f'(1) = 2 \][/tex]
[tex]\[ g'(1) = 6 \][/tex]

Now, we apply these values to the product rule formula:

[tex]\[ h'(1) = f'(1) \cdot g(1) + f(1) \cdot g'(1) \][/tex]

Substitute the values into the formula:

[tex]\[ h'(1) = 2 \cdot 4 + (-5) \cdot 6 \][/tex]

Calculate each term:

[tex]\[ 2 \cdot 4 = 8 \][/tex]
[tex]\[ -5 \cdot 6 = -30 \][/tex]

Add the results:

[tex]\[ h'(1) = 8 + (-30) = 8 - 30 = -22 \][/tex]

Therefore, the value of [tex]\( h'(1) \)[/tex] is:

[tex]\[ h'(1) = -22 \][/tex]

So, the detailed solution shows that [tex]\( h'(1) = -22 \)[/tex].

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