To find the linearization $L(x)$ of the function $f(x) = \sqrt[3]{\sqrt{x}}$ at $a = 125$, we use the following formula for linearization:

$L(x) = f(a) + f'(a)(x - a)$

Step 1: Find $f(a)$

First, we evaluate the function at $a = 125$.

[ f(125) = \sqrt[3]{\sqrt{125}} = \sqrt[3]{\sqrt{125}} = \sqrt[3]{11.18} \approx 2.234 ]

Step 2: Find $f'(x)$

Now we need to compute the derivative $f'(x)$. The function $f(x) = \sqrt[3]{x^{1/2}}$ can be written as:

$f(x) = x^{1/6}$

Using the power rule to differentiate:

$f'(x) = \frac{1}{6} x^{-5/6}$

Step 3: Evaluate $f'(a)$

Now, evaluate the derivative at $a = 125$:

$f'(125) = \frac{1}{6} \cdot 125^{-5/6}$

Since $125 = 5^3$, we can simplify $125^{-5/6}$:

$125^{-5/6} = (5^3)^{-5/6} = 5^{-5/2} = \frac{1}{5^{5/2}} = \frac{1}{55.9} \approx 0.01788$

Thus:

$f'(125) = \frac{1}{6} \times 0.01788 \approx 0.00298$

Step 4: Write the linearization $L(x)$

Now we can plug the values into the linearization formula:

$L(x) = f(125) + f'(125)(x - 125)$

$L(x) \approx 2.234 + 0.00298(x - 125)$

So the linearization is approximately:

$L(x) = 2.234 + 0.00298(x - 125)$

This is the linear approximation of $f(x)$ at $a = 125$.

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Would you like more details on any of the steps?

Here are 5 related questions to explore:

1. How does the linearization approximation change for different values of $a$?
2. What are some practical applications of linearization in real-world problems?
3. How do you find the second derivative of $f(x) = x^{1/6}$?
4. How accurate is the linear approximation near $a = 125$?
5. How can we improve the approximation using higher-order terms?

Tip: Linearization provides a quick way to approximate function values near a point without needing to compute the full function.