find the linearization ( l(x) ) at ( x = a ).
( f(x)=2 x^{3}+3 x + 1 quad a = 2 )
( l(x)=square )

Question

Accepted Answer

# Explanation:
## Step1: Calculate $ f(a) $
Substitute $ x = a = 2 $ into $ f(x)=2x^{3}+3x + 1 $.
$$
\begin{align*}
f(2)&=2\times(2)^{3}+3\times(2)+1\
&=2\times8 + 6+1\
&=16+6 + 1\
&=23
\end{align*}
$$
## Step2: Find the derivative $ f^{\prime}(x) $
Differentiate $ f(x)=2x^{3}+3x + 1 $ using the power rule $ (x^{n})^\prime=nx^{n - 1} $.
$ f^{\prime}(x)=6x^{2}+3 $
## Step3: Calculate $ f^{\prime}(a) $
Substitute $ x = a = 2 $ into $ f^{\prime}(x) $.
$$
\begin{align*}
f^{\prime}(2)&=6\times(2)^{2}+3\
&=6\times4+3\
&=24 + 3\
&=27
\end{align*}
$$
## Step4: Use the linearization formula $ L(x)=f(a)+f^{\prime}(a)(x - a) $
Substitute $ f(2) = 23 $, $ f^{\prime}(2)=27 $, and $ a = 2 $ into the formula.
$$
\begin{align*}
L(x)&=23+27(x - 2)\
&=23+27x-54\
&=27x-31
\end{align*}
$$

# Answer:
$ L(x)=27x - 31 $

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate \( f(a) \)

Step2: Find the derivative \( f^{\prime}(x) \)

Step3: Calculate \( f^{\prime}(a) \)

Step4: Use the linearization formula \( L(x)=f(a)+f^{\prime}(a)(x - a) \)

Answer: