Let's go through each part of this problem to confirm the solution.

Part (a): Find $f'(0)$

Given: $f(x) = (1 + 4x)^{\frac{1}{3}}$

To find $f'(x)$, we'll use the chain rule.

1. Differentiate $f(x) = (1 + 4x)^{\frac{1}{3}}$: $f'(x) = \frac{1}{3}(1 + 4x)^{-\frac{2}{3}} \cdot 4$ Simplifying, we get: $f'(x) = \frac{4}{3} (1 + 4x)^{-\frac{2}{3}}$

2. Evaluate at $x = 0$: $f'(0) = \frac{4}{3} (1 + 4 \cdot 0)^{-\frac{2}{3}} = \frac{4}{3}$

So, $f'(0) = \frac{4}{3}$.

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Part (b): Find the linearization $L(x)$ of $f(x)$ at $x = 0$

The linearization of $f(x)$ at $x = 0$ is given by: $L(x) = f(0) + f'(0) \cdot (x - 0)$

1. Calculate $f(0)$: $f(0) = (1 + 4 \cdot 0)^{\frac{1}{3}} = 1$

2. Substitute values into $L(x)$: $L(x) = 1 + \frac{4}{3} \cdot x$

So, $L(x) = 1 + \frac{4}{3}x$.

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Part (c): Find $x$ such that $f(x) = \left(\frac{9}{10}\right)^{\frac{1}{3}}$

We need to find $x$ such that: $f(x) = (1 + 4x)^{\frac{1}{3}} = \left(\frac{9}{10}\right)^{\frac{1}{3}}$

1. Rewrite in terms of powers: $1 + 4x = \frac{9}{10}$

2. Solve for $x$: $4x = \frac{9}{10} - 1 = -\frac{1}{10}$ $x = -\frac{1}{40} = -0.025$

So, $x = -0.025$.

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Part (d): Use linearization to estimate $\left(\frac{9}{10}\right)^{\frac{1}{3}}$

Using the linear approximation $L(x) = 1 + \frac{4}{3} x$ with $x = -0.025$: $L(-0.025) = 1 + \frac{4}{3} \cdot (-0.025)$ $= 1 - \frac{4}{3} \cdot 0.025$ $= 1 - \frac{0.1}{3}$ $= 1 - 0.0333 \approx 0.9667$

Therefore, the estimated value of $\left(\frac{9}{10}\right)^{\frac{1}{3}}$ using the linearization is approximately $0.9667$.

Let me know if you need further details, or if you'd like to check the steps with different values. Here are some related questions:

1. How is the chain rule used to find the derivative in functions involving exponents?
2. Why is linearization useful for approximations near a certain point?
3. What other methods can approximate $\left(\frac{9}{10}\right)^{\frac{1}{3}}$?
4. How does the accuracy of the linear approximation vary with distance from $x = 0$?
5. Can we improve this approximation with higher-order terms?

Tip: Linearization is especially useful when evaluating functions at points near the point of expansion, as the function behaves similarly to the tangent line at that location.