Answer
The absolute value of the difference in the two results: $0.0792$
Work Step by Step
$f(x) = x^{\frac{1}{2}}$
$f'(x)=\frac{1}{2}x^{\frac{-1}{2}}$
$f(x + \Delta x) \approx f(x) + f'(x)dx=\sqrt x + \frac{1}{2\sqrt x}dx$
$f(23)=f(16+7) \approx \sqrt 16 + \frac{1}{2\sqrt 16}(7)=4+\frac{7}{8}=4\frac{7}{8}\approx 4.875$
The true value of this, found using a calculator, is $\sqrt 23 \approx 4.7958$
The absolute value of the difference in the two results: $0.0792$
