Answer
The absolute value of the difference in the two results: $0.0020$
Work Step by Step
$f(x) = x^{\frac{1}{2}}$
$f'(x)=\frac{1}{2}x^{\frac{-1}{2}}$
$f(x + \Delta x) \approx f(x) + f'(x)dx=\sqrt x + \frac{1}{2\sqrt x}dx$
$f(17.02)=f(16+1.02) \approx \sqrt 16 + \frac{1}{2\sqrt 16}(1.02)= 4 +\frac{51}{400} \approx 4.1275$
The true value of this, found using a calculator, is $\sqrt 23 \approx 4.1255$
The absolute value of the difference in the two results: $0.0020$