4. (5 points) consider ( f(x)=ln (1 + 3x) ) for ( x ) near 2.\na. (3 points) find the local linearization of…

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.