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Suppose \( f(x) \) and \( f^{\prime}(x) \) have the values shown.

\begin{tabular}{|c|r|r|r|r|r|}
\hline
[tex]$x$[/tex] & 44 & 45 & 46 & 47 & 48 \\
\hline
[tex]$f(x)$[/tex] & -6 & 8 & -2 & 1 & -3 \\
\hline
[tex]$f^{\prime}(x)$[/tex] & -5 & 8 & 4 & 1 & -10 \\
\hline
\end{tabular}

Let \( g(x) = f\left(5 x^2 - 4 x - 61\right) \).

Find \( g^{\prime}(5) \).

[tex]\( g^{\prime}(5) = \square \)[/tex]



Answer :

GinnyAnswer
To find the value of \( g'(5) \), we start by understanding the function \( g(x) \) and then compute its derivative.

Given:
[tex]\[ g(x) = f(5x^2 - 4x - 61) \][/tex]

To find \( g'(x) \), we need to apply the chain rule. The chain rule for derivatives tells us:
[tex]\[ g'(x) = f'(u) \cdot \frac{du}{dx} \][/tex]
where \( u = 5x^2 - 4x - 61 \).

First, let's differentiate \( u \) with respect to \( x \):
[tex]\[ u = 5x^2 - 4x - 61 \][/tex]
[tex]\[ \frac{du}{dx} = 10x - 4 \][/tex]

Next, we plug this into the chain rule formula:
[tex]\[ g'(x) = f'(5x^2 - 4x - 61) \cdot (10x - 4) \][/tex]

We now need to evaluate this expression at \( x = 5 \):
[tex]\[ g'(5) = f'(5(5^2) - 4(5) - 61) \cdot (10(5) - 4) \][/tex]

First, compute the argument of \( f' \) (i.e., the inner function):
[tex]\[ 5(5^2) - 4(5) - 61 = 5(25) - 20 - 61 = 125 - 20 - 61 = 44 \][/tex]

We now look up \( f'(44) \) in the given table:
[tex]\[ f'(44) = -5 \][/tex]

Using this value in our derivative expression:
[tex]\[ g'(5) = f'(44) \cdot (10(5) - 4) = -5 \cdot (50 - 4) = -5 \cdot 46 = -230 \][/tex]

Therefore:
[tex]\[ g'(5) = -230 \][/tex]

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