4. (5 points) consider ( f(x)=ln (1 + 3x) ) for ( x ) near 2.
a. (3 points) find the local linearization of ( f(x) ) at ( x = 2 ).
b. (2 points) is the local linearization an under - or over - estimate? use the second derivative to justify your answer.

Question

Accepted Answer

# Explanation:
## Step1: Calculate $f(2)$
$$
\begin{align*}
f(x)&=\ln(1 + 3x)\
f(2)&=\ln(1+3\times2)=\ln(7)
\end{align*}
$$
## Step2: Calculate $f^{\prime}(x)$ and $f^{\prime}(2)$
$$
\begin{align*}
f^{\prime}(x)&=\frac{3}{1 + 3x}\
f^{\prime}(2)&=\frac{3}{1+3\times2}=\frac{3}{7}
\end{align*}
$$
## Step3: Find the local linearization $L(x)$
The formula for local linearization is $L(x)=f(a)+f^{\prime}(a)(x - a)$, where $a = 2$
$$
\begin{align*}
L(x)&=\ln(7)+\frac{3}{7}(x - 2)\
&=\frac{3}{7}x+\ln(7)-\frac{6}{7}
\end{align*}
$$
## Step4: Calculate $f^{\prime\prime}(x)$
$$
\begin{align*}
f^{\prime}(x)&=\frac{3}{1 + 3x}=3(1 + 3x)^{-1}\
f^{\prime\prime}(x)&=- 9(1 + 3x)^{-2}=-\frac{9}{(1 + 3x)^{2}}
\end{align*}
$$
Since $f^{\prime\prime}(x)<0$ for all $x$ in the domain of $f(x)$ (where $1+3x>0$), the function $y = f(x)$ is concave - down.

# Answer:
a. The local linearization $L(x)=\frac{3}{7}x+\ln(7)-\frac{6}{7}$
b. The local linearization is an over - estimate. Because $f^{\prime\prime}(x)=-\frac{9}{(1 + 3x)^{2}}<0$, the function $y = f(x)$ is concave - down. By the property of concave - down functions, the tangent line (local linearization) lies above the function curve.

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate \(f(2)\)

Step2: Calculate \(f^{\prime}(x)\) and \(f^{\prime}(2)\)

Step3: Find the local linearization \(L(x)\)

Step4: Calculate \(f^{\prime\prime}(x)\)

Answer: