SOLUTION 1: Begin with the function
$$ f(x)= \sqrt{x} $$
and choose
$$ x-values: 25 \rightarrow 28 $$
so that
$$ \Delta x = 28-25 = 3 $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ 1 \over 2 }x^{-1/2} = \displaystyle{ 1 \over 2 \sqrt{x} } $$
The exact change of $y-$values is
$$ \Delta y = f(28) - f(25) $$
$$ = \sqrt{28} - \sqrt{25} $$
$$ = \sqrt{28} - 5 $$
The Differential is
$$ dy = f'(25) \ \Delta x $$
$$ = \displaystyle{ 1 \over 2 \sqrt{25} } \cdot (3) $$
$$ = \displaystyle{ 1 \over 2 (5) } (3) $$
$$ = \displaystyle{ 1 \over 10 } (3) $$
$$ = 0.3 $$
We will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \sqrt{28} - 5 \approx 0.3 \ \ \ \ \longrightarrow $$
$$ \sqrt{28} \approx 5 + 0.3 \ \ \ \ \longrightarrow $$
$$ \sqrt{28} \approx 5.3 $$
NOTE: The number 25 was chosen for its proximity to 28 and for it's convenient square root. Check the accuracy of the final estimate using a CALCULATOR: $ \sqrt{28} \approx 5.2915 $
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