Answer is: option3
\( 2 + \frac{h}{12} \)Solution:
Let:
\[ f(x) = \sqrt[3]{x} \]
We expand \( f(x) \) around \( x = 8 \), so we define:
\[ f(x) = x^{1/3} \]
The derivative is:
\[ f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} \]
Evaluating at \( x = 8 \):
\[ f(8) = \sqrt[3]{8} = 2 \]
\[ f'(8) = \frac{1}{3\sqrt[3]{8^2}} = \frac{1}{3 \cdot 4} = \frac{1}{12} \]
Using the linear approximation:
\[ f(8 + h) \approx f(8) + f'(8) \cdot h \]
\[ \sqrt[3]{8 + h} \approx 2 + \frac{h}{12} \]