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To solve the problem of finding the linearization \( L(x) \) of the function \( f(x) = \frac{5e^x}{x^4} \) at \( x = 1 \), let’s go through the linearization process step-by-step.

### Step 1: Understand Linearization
The linearization \( L(x) \) of a function \( f(x) \) at \( x = a \) is given by:
\[
L(x) = f(a) + f'(a)(x - a)
\]
where \( f(a) \) is the value of the function at \( x = a \), and \( f'(a) \) is the derivative of \( f(x) \) evaluated at \( x = a \).

### Step 2: Calculate \( f(1) \)
The function is:
\[
f(x) = \frac{5e^x}{x^4}
\]
Substitute \( x = 1 \):
\[
f(1) = \frac{5e^1}{1^4} = 5e
\]

### Step 3: Calculate \( f'(x) \)
To find \( f'(x) \), we’ll need to use the quotient rule. For \( f(x) = \frac{u(x)}{v(x)} \) with \( u(x) = 5e^x \) and \( v(x) = x^4 \), the derivative \( f'(x) \) is:
\[
f'(x) = \frac{v(x) u'(x) - u(x) v'(x)}{(v(x))^2}
\]
where:
- \( u'(x) = 5e^x \)
- \( v'(x) = 4x^3 \)

So,
\[
f'(x) = \frac{x^4 \cdot 5e^x - 5e^x \cdot 4x^3}{(x^4)^2}
\]
Simplify this expression:
\[
f'(x) = \frac{5e^x (x^4 - 4x^3)}{x^8} = \frac{5e^x (x - 4)}{x^5}
\]

### Step 4: Evaluate \( f'(1) \)
Now, substitute \( x = 1 \) into \( f'(x) \):
\[
f'(1) = \frac{5e^1 (1 - 4)}{1^5} = \frac{5e(-3)}{1} = -15e
\]

### Step 5: Write the Linearization \( L(x) \)
Using the formula \( L(x) = f(1) + f'(1)(x - 1) \), we get:
\[
L(x) = 5e + (-15e)(x - 1)
\]
Distribute the terms:
\[
L(x) = 5e - 15e(x - 1) = 5e - 15ex + 15e
\]
Combine constants:
\[
L(x) = 20e - 15ex
\]

### Final Answer
\[
L(x) = 20e - 15ex
\]

Let me know if you want further clarification or if you have more questions!

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Here are five questions related to this process for deeper understanding:
1. How would the linearization change if \( f(x) \) had a different exponent in the denominator?
2. What if we were asked to find the linearization at a different point, say \( x = 2 \)?
3. Why is the quotient rule necessary for finding \( f'(x) \) in this problem?
4. Can the linearization approximation be used for values of \( x \) far from 1? Why or why not?
5. How would the linearization differ if we applied it to a function that’s not smooth at \( x = 1 \)?

**Tip:** Remember that the linearization provides a good approximation only near the point of tangency. For values farther away, the approximation may become less accurate.

Calculate the linearization L of f(x) = 5e^x / x^4 at x = 1. Express your answer in terms of e and x.

This problem demonstrates how to find the linearization L(x) of the function f(x) = 5e^x / x^4 at x = 1. The solution includes applying the linear approximation formula and using the quotient rule for derivatives. Suitable for advanced high school or college calculus students.

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