## Calculus with Applications (10th Edition)

The absolute value of the difference in the two results: $0.0001$
$f(x) = x^{\frac{1}{2}}$ $f'(x)=\frac{1}{2}x^{\frac{-1}{2}}$ $f(x + \Delta x) \approx f(x) + f'(x)dx=\sqrt x + \frac{1}{2\sqrt x}dx$ $f(145)=f(144+1) \approx \sqrt 144 + \frac{1}{2\sqrt 144}(1)=12+\frac{1}{24}=12\frac{1}{24}\approx12.0417$ 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: $0.0001$