Using linear approximation to find \( \sqrt{4.7} \) Use \( y=\sqrt{3-x} ; x_{0}=-1 \quad(2 d p) \quad\left[\begin{array}{lll}3\end{array}\right] \)

To find \( \sqrt{4.7} \) using linear approximation, we start with the function \( y = \sqrt{3 - x} \). We want to approximate at \( x = -1 \) (since \( 4.7 = 3 - x \), this means \( x = -1.7 \)). We first find the derivative \( y' = -\frac{1}{2\sqrt{3-x}} \). Calculate \( y \) and \( y' \) at \( x_0 = -1 \): - \( y(-1) = \sqrt{3 - (-1)} = \sqrt{4} = 2 \) - \( y'(-1) = -\frac{1}{2\sqrt{4}} = -\frac{1}{4} \) Using the linear approximation formula: \[ y \approx y(x_0) + y'(x_0)(x - x_0) \] We plug in \( x = -1.7 \) and \( x_0 = -1 \): \[ y \approx 2 - \frac{1}{4}(-1.7 + 1) = 2 - \frac{1}{4}(-0.7) = 2 + 0.175 = 2.175 \] Thus, \( \sqrt{4.7} \approx 2.175 \). Don't forget, this approximation gives you a close estimate and shows the brilliant application of calculus in predicting values that seem out of reach!