Find a linearization at a suitably chosen integer near a at which the given function and its derivative are easy to evaluate.
-$$f ( x ) = \sqrt { x } , a = 9.1$$
A) $L ( x ) = \frac { 3 } { 2 } - \frac { 1 } { 6 } x$
B) $L ( x ) = \frac { 3 } { 2 } + \frac { 1 } { 3 } x$
C) $L ( x ) = \frac { 1 } { 2 } + \frac { 1 } { 2 } x$
D) $L ( x ) = \frac { 3 } { 2 } + \frac { 1 } { 6 } x$

Question

Quizplus · Accepted Answer

The linearization of $f(x) = \sqrt{x}$ at a point close to $a = 9.1$ is best done at $x = 9$ because the square root and its derivative are easy to calculate at this point. The formula for linearization is $L(x) = f(a) + f'(a)(x - a)$. At $x = 9$, $f(9) = \sqrt{9} = 3$ and $f'(x) = \frac{1}{2\sqrt{x}}$, so $f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}$. Substituting these values into the linearization formula gives $L(x) = 3 + \frac{1}{6}(x - 9) = 3 + \frac{1}{6}x - \frac{3}{2} = \frac{3}{2} + \frac{1}{6}x$, which matches option D.

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