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!

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.