Answer
$6.00248125$
Work Step by Step
The local quadratic approximation can be calculated as:
$f(x)=f(x_0) +f^{\prime}(x_0)(x-x_0)+f^{\prime\prime}(x_0)(x-x_0)^2$
Now, the local quadratic approximation of $f$ at $x_0=36$ is:
$ \sqrt {36.03} \approx f(36.03) \approx 6+\dfrac{36.03-36}{12}-\dfrac{(36.03-36)^2}{48} \\ \approx 6.00248125$
By using a calculator, we have: $ \sqrt {36.03} \approx 6.00248125$
Now, the approximated value up to 4 decimals is $6.00248125$.
