Answer
The absolute value of the difference in the two results: $0.0001$
Work Step by Step
$f(x) = x^{\frac{1}{2}}$
$f'(x)=\frac{1}{2}x^{\frac{-1}{2}}$
$f(x + \Delta x) \approx f(x) + f'(x)dx=\sqrt x + \frac{1}{2\sqrt x}dx$
$f(145)=f(144+1) \approx \sqrt 144 + \frac{1}{2\sqrt 144}(1)=12+\frac{1}{24}=12\frac{1}{24}\approx12.0417$
The true value of this, found using a calculator, is $\sqrt 145 \approx 12.0416$
The absolute value of the difference in the two results: $0.0001$
