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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].

\begin{tabular}{|c|c|c|c|}
\hline
[tex][tex]$x$[/tex][/tex] & [tex][tex]$f(x)$[/tex][/tex] & [tex][tex]$g(x)$[/tex][/tex] & [tex][tex]$f^{\prime}(x)$[/tex][/tex] & [tex][tex]$g^{\prime}(x)$[/tex][/tex] \\
\hline
-4 & -9 & 8 & 6 & -2 \\
\hline
-1 & -4 & 6 & 3 & 1 \\
\hline
3 & 1 & -3 & 2 & 4 \\
\hline
1 & -8 & 6 & 6 & -2 \\
\hline
6 & 3 & 7 & -1 & -8 \\
\hline
\end{tabular}

Use the values in the table to find [tex][tex]$h^{\prime}(-1)$[/tex][/tex], where
[tex]\[
h(x) = \frac{3}{[g(x)]^4}
\][/tex]
(round decimal answer to 3 decimal places if needed)



Answer :

GinnyAnswer
To find [tex]\(h'(-1)\)[/tex] where [tex]\(h(x) = \frac{3}{[g(x)]^4}\)[/tex], let's follow these steps:

1. Express [tex]\(h(x)\)[/tex] explicitly:
[tex]\[ h(x) = 3 [g(x)]^{-4} \][/tex]

2. Compute the derivative [tex]\(h'(x)\)[/tex] using the chain rule:
[tex]\[ h(x) = 3 [g(x)]^{-4} \][/tex]
Let [tex]\(u = g(x)\)[/tex]. Then [tex]\(h(x) = 3 u^{-4}\)[/tex].

Compute the derivative [tex]\(\frac{d}{dx}[3 u^{-4}]\)[/tex]:
[tex]\[ \frac{d}{dx}[3 u^{-4}] = 3 \cdot (-4) u^{-5} \frac{du}{dx} \][/tex]
[tex]\[ \frac{d}{dx}[3 u^{-4}] = -12 u^{-5} \frac{du}{dx} \][/tex]

3. Substitute back [tex]\(u = g(x)\)[/tex] and the chain rule result:
[tex]\[ h'(x) = -12 [g(x)]^{-5} g'(x) \][/tex]

4. Evaluate [tex]\(h'(-1)\)[/tex] using the given values from the table:
From the table:
[tex]\[ g(-1) = 6 \][/tex]
[tex]\[ g'(-1) = 1 \][/tex]

Substitute [tex]\(g(-1)\)[/tex] and [tex]\(g'(-1)\)[/tex] into the derivative formula:
[tex]\[ h'(-1) = -12 [g(-1)]^{-5} g'(-1) \][/tex]

5. Calculate [tex]\(h'(-1)\)[/tex]:
[tex]\[ h'(-1) = -12 [6]^{-5} \cdot 1 \][/tex]
[tex]\[ h'(-1) = -12 \cdot 6^{-5} \cdot 1 \][/tex]

6. Simplify [tex]\(6^{-5}\)[/tex]:
[tex]\[ 6^{-5} = \frac{1}{6^5} = \frac{1}{7776} \][/tex]
[tex]\[ h'(-1) = -12 \cdot \frac{1}{7776} \][/tex]
[tex]\[ h'(-1) = -\frac{12}{7776} = -\frac{1}{648} \approx -0.001543 \][/tex]

7. Round the answer to 3 decimal places:
[tex]\[ h'(-1) \approx -0.002 \][/tex]

So, the value of [tex]\(h'(-1)\)[/tex] is [tex]\(-0.002\)[/tex].

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