https://file.aiquickdraw.com/mathbot/file/172739527042776h6thgg.jpg

Let's analyze the problem step by step.

### Part (a) Analysis:

We are given that the tangent line to the curve \( y = f(x) \) at the point \( (1, 13) \) is \( y = 9 + 4x \). Using Newton's method, we are asked to locate a root of the equation \( f(x) = 0 \), with an initial approximation \( x_1 = 1 \). We need to find the second approximation \( x_2 \).

- The equation of the tangent line gives us two key pieces of information:
  - The point \( (1, 13) \) is on the curve, so \( f(1) = 13 \).
  - The slope of the tangent line is 4, so \( f'(1) = 4 \).
  
  Newton's method is defined by the recurrence relation:
  \[
  x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
  \]
  Given \( x_1 = 1 \), we can calculate \( x_2 \).

Now I'll calculate \( x_2 \) using Newton's method.

### Part (b) Analysis:

We are given an initial approximation \( x_1 = 7 \), and the second approximation \( x_2 = -1 \). Additionally, the tangent line at \( x = 7 \) passes through the point \( (15, 6) \). We are asked to find \( f(7) \).

Using the Newton’s method formula, we can calculate \( f'(7) \) from the given information about the tangent line and the points it passes through.

### Part (c) Analysis:

We need to use Newton's method with the initial approximation \( x_1 = 4 \) to find the second approximation \( x_2 \) to the root of the equation \( x^3 = 5x + 4 \).

I will now solve each part, starting with part (a).For **part (a)**, the second approximation using Newton's method is \( x_2 = -2.25 \).

Let me proceed with the next part, **part (b)**, to find \( f(7) \).### Part (b) Details:

We are given:
- The initial approximation \( x_1 = 7 \)
- The second approximation \( x_2 = -1 \)
- The tangent line to \( f(x) \) at \( x = 7 \) passes through the point \( (15, 6) \).

We need to find the value of \( f(7) \). To do this, let's first determine the slope of the tangent line using the given points.

The slope of the line passing through \( (7, f(7)) \) and \( (15, 6) \) is:
\[
\text{slope} = \frac{6 - f(7)}{15 - 7}
\]
Given that the tangent line represents the derivative at \( x = 7 \), we can relate this to \( f'(7) \).

Next, we will calculate the exact value of \( f(7) \).For **part (b)**, the value of \( f(7) \) is \( \frac{5}{2} \), or \( 2.5 \).

Now, let's move to **part (c)**, where we will use Newton's method to find the second approximation \( x_2 \) for the equation \( x^3 = 5x + 4 \), with an initial approximation \( x_1 = 4 \).

Suppose that the tangent line to the curve y = f(x) at the point (1, 13) has equation y = 9 + 4x. If Newton's method is used to locate a root of the equation f(x) = 0 and the initial approximation is x1 = 1, find the second approximation x2.

Suppose that Newton's method is used to locate a root of the equation f(x) = 0 with initial approximation x1 = 7. If the second approximation is found to be x2 = -1, and the tangent line to f(x) at x = 7 passes through the point (15, 6), find f(7).

Use Newton's method with initial approximation x1 = 4 to find x2, the second approximation to the root of the equation x^3 = 5x + 4.

Explore how to use Newton's method to find roots of equations involving tangent lines and derivatives. Learn to find second approximations and analyze tangent lines at given points. Suitable for advanced high school and college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation