Using differentials,find the approximate valu | vedclass

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Question

(A) We use the formula for linear approximation: $f(x + \Delta x) \approx f(x) + \Delta x \cdot f'(x)$.Let $f(x) = \sqrt{x}$. Then $f'(x) = \frac{1}{2

Vedclass · Accepted Answer

$0.2867$

Vedclass · Answer

(A) We use the formula for linear approximation: $f(x + \Delta x) \approx f(x) + \Delta x \cdot f'(x)$.Let $f(x) = \sqrt{x}$. Then $f'(x) = \frac{1}{2\sqrt{x}}$.We choose $x = 0.09$ because $\sqrt{0.09} = 0.3$ is a perfect square root.Then $\Delta x = 0.082 - 0.09 = -0.008$.Substituting these values into the formula:$f(0.082) \approx f(0.09) + (-0.008) \cdot f'(0.09)$$f(0.082) \approx 0.3 + (-0.008) \cdot \left(\frac{1}{2 \cdot 0.3}\right)$$f(0.082) \approx 0.3 - \frac{0.008}{0.6}$$f(0.082) \approx 0.3 - 0.01333...$$f(0.082) \approx 0.28666... \approx 0.2867$.

Using differentials,find the approximate value of $\sqrt{0.082}$. Let $f(x) = \sqrt{x}$.

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