Let \( f(x)=\sqrt{x+1} \) and use the Linear Approximation to this function at \( a=3 \) with \( \Delta x=0.5 \) to estimate \( f(3.5)-f(3)=\Delta f \approx d f=\square \) Preview My Answers Submit Answers
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To estimate \( f(3.5) - f(3) \) using linear approximation, we'll first need to find \( f'(x) \) at \( a = 3 \). The derivative of \( f(x) = \sqrt{x+1} \) is \( f'(x) = \frac{1}{2\sqrt{x+1}} \). Evaluating this at \( a = 3 \): \[ f'(3) = \frac{1}{2\sqrt{3+1}} = \frac{1}{2\sqrt{4}} = \frac{1}{4} \] Now, we can use \( f(3) = \sqrt{3+1} = \sqrt{4} = 2 \) and \( \Delta x = 0.5 \) to find the approximate change: \[ \Delta f \approx d f = f'(3) \Delta x = \frac{1}{4} \times 0.5 = \frac{1}{8} = 0.125 \] So, \( f(3.5) - f(3) \approx 0.125 \).