Answer
(a) $f\left( x \right) \approx 1 + \dfrac{1}{2}\left( {x - 1} \right) - \dfrac{1}{8}{\left( {x - 1} \right)^2}$
(b) $\sqrt {1.1} \approx 1.04875$
Calculator: $\sqrt {1.1} \approx 1.04881$
Work Step by Step
(a) Let $f\left( x \right) = \sqrt x $. Then, the derivatives are
$f'\left( x \right) = \dfrac{1}{{2\sqrt x }}$, ${\ \ \ \ \ }$ $f{\rm{''}}\left( x \right) = - \dfrac{1}{{4{x^{3/2}}}}$
The local quadratic approximation of $f$ at ${x_0} = 1$ is
$f\left( x \right) \approx f\left( 1 \right) + f'\left( 1 \right)\left( {x - 1} \right) + \dfrac{{f{\rm{''}}\left( 1 \right)}}{{2!}}{\left( {x - 1} \right)^2}$
$f\left( x \right) \approx 1 + \dfrac{1}{2}\left( {x - 1} \right) - \dfrac{1}{8}{\left( {x - 1} \right)^2}$
(b) With $x = 1.1$, we have
$f\left( {1.1} \right) = \sqrt {1.1} \approx 1 + \dfrac{1}{2}\left( {1.1 - 1} \right) - \dfrac{1}{8}{\left( {1.1 - 1} \right)^2}$
$\sqrt {1.1} \approx 1.04875$
Using a calculating facility, we obtain $\sqrt {1.1} \approx 1.04881$.
