Use linear approximation, i.e. the tangent line, to approximate \( \sqrt[3]{27.3} \) as follows: Let \( f(x)=\sqrt[3]{x} \). The equation of the tangent line to \( f(x) \) at \( x=27 \) can be written in the form \( y=m x+b \) where \( m \) is: and where \( b \) is: Using this, we find our approximation for \( \sqrt[3]{27.3} \) is

To find the linear approximation of \( f(x) = \sqrt[3]{x} \) at \( x = 27 \), we first need the slope of the tangent line, which is given by the derivative \( f'(x) \). Calculating \( f'(x) = \frac{1}{3} x^{-\frac{2}{3}} \) and then evaluating it at \( x = 27 \) gives us \( f'(27) = \frac{1}{3} \times 27^{-\frac{2}{3}} = \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} \). Therefore, \( m = \frac{1}{27} \). Next, we find \( b \) using the point \( (27, f(27)) = (27, 3) \). The equation of the tangent line is \( y - 3 = \frac{1}{27}(x - 27) \). Rearranging, we find \( y = \frac{1}{27}x + 2 \). Plugging in \( x = 27.3 \) gives us \( y \approx \frac{1}{27}(27.3) + 2 \approx 3.010 \). Thus, our approximation for \( \sqrt[3]{27.3} \) is approximately \( 3.010 \).