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 $x_1 = 4$.