Use the Taylor series to linearize the following (show all your steps):
3.1
 (10)
𝑓𝑓(𝑥𝑥) = 5𝑥𝑥2 − 2𝑥𝑥, 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑥𝑥 = 3
3.2 (10)
𝑓𝑓(𝑥𝑥) = −3𝑥𝑥3 + 30𝑥𝑥, 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑥𝑥 = −3

To linearize a function \( f(x) \) at a given point \( x_0 \), we use the first two terms of its Taylor series expansion around \( x_0 \), which can be written as:

\[
f(x) \approx f(x_0) + f'(x_0)(x - x_0)
\]

This approximation is the linearization of the function at \( x_0 \). We need to calculate \( f(x_0) \) and \( f'(x_0) \), which involves computing the first derivative of \( f(x) \) and evaluating both \( f(x) \) and \( f'(x) \) at the given point \( x_0 \).

### 3.1: Linearizing \( f(x) = 5x^2 - 2x \) at \( x = 3 \)

**Step 1: Find the first derivative of \( f(x) \)**

\[
f(x) = 5x^2 - 2x
\]

Differentiate \( f(x) \) with respect to \( x \):

\[
f'(x) = \frac{d}{dx}(5x^2 - 2x) = 10x - 2
\]

**Step 2: Evaluate \( f(x_0) \) and \( f'(x_0) \) at \( x_0 = 3 \)**

\[
f(3) = 5(3)^2 - 2(3) = 5(9) - 6 = 45 - 6 = 39
\]

\[
f'(3) = 10(3) - 2 = 30 - 2 = 28
\]

**Step 3: Write the linearized approximation**

Using the linearization formula \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \), we substitute \( x_0 = 3 \), \( f(3) = 39 \), and \( f'(3) = 28 \):

\[
f(x) \approx 39 + 28(x - 3)
\]

This is the linearized form of \( f(x) \) at \( x = 3 \).

### 3.2: Linearizing \( f(x) = -3x^3 + 30x \) at \( x = -3 \)

**Step 1: Find the first derivative of \( f(x) \)**

\[
f(x) = -3x^3 + 30x
\]

Differentiate \( f(x) \) with respect to \( x \):

\[
f'(x) = \frac{d}{dx}(-3x^3 + 30x) = -9x^2 + 30
\]

**Step 2: Evaluate \( f(x_0) \) and \( f'(x_0) \) at \( x_0 = -3 \)**

\[
f(-3) = -3(-3)^3 + 30(-3) = -3(-27) - 90 = 81 - 90 = -9
\]

\[
f'(-3) = -9(-3)^2 + 30 = -9(9) + 30 = -81 + 30 = -51
\]

**Step 3: Write the linearized approximation**

Using the linearization formula \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \), we substitute \( x_0 = -3 \), \( f(-3) = -9 \), and \( f'(-3) = -51 \):

\[
f(x) \approx -9 - 51(x + 3)
\]

This is the linearized form of \( f(x) \) at \( x = -3 \).

### Summary:
- **For 3.1**: The linearization of \( f(x) = 5x^2 - 2x \) at \( x = 3 \) is:
  \[
  f(x) \approx 39 + 28(x - 3)
  \]
- **For 3.2**: The linearization of \( f(x) = -3x^3 + 30x \) at \( x = -3 \) is:
  \[
  f(x) \approx -9 - 51(x + 3)
  \]

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Would you like more details on any of the steps?

Here are 5 additional questions that relate to linearization:
1. How does the accuracy of linearization change as you move further from the point of approximation?
2. How would the linearization formula change if we included the second derivative?
3. Can you explain the geometric interpretation of linearization?
4. How is linearization used in solving differential equations?
5. What are some real-world applications of linearizing nonlinear functions?

**Tip:** Linearization is a powerful tool for simplifying complex functions, especially when analyzing behavior near a specific point. Always ensure the function is differentiable at the point of approximation!

Use the Taylor series to linearize the following (show all your steps):

3.1: f(x) = 5x^2 - 2x, where x = 3
3.2: f(x) = -3x^3 + 30x, where x = -3

Learn how to linearize the functions f(x) = 5x^2 - 2x at x = 3 and f(x) = -3x^3 + 30x at x = -3 using the Taylor series. This solution covers key calculus concepts such as derivatives and linear approximation. Suitable for undergraduate students studying Calculus I or II.

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