To find the linearization of the function $f(x) = \sqrt[3]{7 + x}$ centered at $a = 1$, we first need to calculate $f(a)$ and $f'(a)$.

1. Calculate $f(1)$: [ f(1) = \sqrt[3]{7 + 1} = \sqrt[3]{8} = 2 ]

2. Calculate the derivative $f'(x)$: Using the chain rule: $f'(x) = \frac{1}{3}(7 + x)^{-\frac{2}{3}} \cdot (1) = \frac{1}{3\sqrt[3]{(7+x)^2}}$

3. Evaluate $f'(1)$: [ f'(1) = \frac{1}{3\sqrt[3]{(7 + 1)^2}} = \frac{1}{3\sqrt[3]{8^2}} = \frac{1}{3 \cdot 4} = \frac{1}{12} ]

4. Construct the linearization $L(x)$: The formula for the linear approximation (linearization) at $a$ is: $L(x) = f(a) + f'(a)(x - a)$ Substituting the values we found: $L(x) = 2 + \frac{1}{12}(x - 1)$

So, the linearization is: $\boxed{L(x) = 2 + \frac{1}{12}(x - 1)}$

Use the linearization to approximate $\sqrt[3]{8.06}$:

To approximate $\sqrt[3]{8.06}$, we can use the linearization at $x = 1$:

1. Substitute $x = 8.06$ into $L(x)$: $L(8.06) = 2 + \frac{1}{12}(8.06 - 1)$ $= 2 + \frac{1}{12}(7.06) = 2 + \frac{7.06}{12}$ $= 2 + 0.5883 \approx 2.5883$

Thus, the approximation for $\sqrt[3]{8.06}$ is: $\boxed{2.5883}$

Would you like any details or have any questions? Here are some related questions:

1. What is the significance of linearization in calculus?
2. How does the linear approximation compare to the actual value?
3. Can you find the linearization at a different point, say $a = 2$?
4. What would be the effect of using a point further away from $a = 1$ for linearization?
5. How do you find the derivative of more complex functions?

Tip: When approximating values, linearization is most accurate for points close to the center of approximation.