Find a linearization at a suitably chosen integer near a at which the given function and its derivative are easy to evaluate.
-$$f ( x ) = \frac { x } { x + 1 } , a = 5.9$$
A) $L ( x ) = \frac { 1 } { 49 } + \frac { 36 } { 49 } x$
B) $L ( x ) = \frac { 36 } { 49 } + \frac { 1 } { 49 } x$
C) $L ( x ) = \frac { 1 } { 49 } + \frac { 6 } { 49 } x$
D) $L ( x ) = \frac { 6 } { 49 } + \frac { 1 } { 49 } x$

Question

Quizplus · Accepted Answer

The linearization of a function $f(x)$ at a point $a$ is given by $L(x) = f(a) + f'(a)(x - a)$. For $f(x) = \frac{x}{x+1}$, we choose $a = 6$ because it's the integer closest to $5.9$. At $x = 6$, $f(6) = \frac{6}{7} = \frac{36}{42}$. The derivative $f'(x) = \frac{1}{(x+1)^2}$, so $f'(6) = \frac{1}{49}$. Plugging $a = 6$ and these values into the linearization formula, we get $L(x) = \frac{36}{42} + \frac{1}{49}(x - 6)$. Simplifying, $L(x) = \frac{36}{42} + \frac{1}{49}x - \frac{6}{49}$, which simplifies further to $L(x) = \frac{36}{49} + \frac{1}{49}x$, matching choice B.

Find a Linearization at a Suitably Chosen Integer near a at Which