The image asks for solving the approximation of the root $r$ of the equation $f(x) = x^3 - 5x + 4 = 0$ using Newton's method. Given the starting point $x_0 = 1$, we need to calculate the tangent line and the approximation $x_1$.

To solve this step-by-step, we can follow Newton's method, which involves the formula:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Step 1: Calculate $f(x)$ and $f'(x)$

Given $f(x) = x^3 - 5x + 4$, the derivative $f'(x)$ is:

$f'(x) = 3x^2 - 5$

Step 2: Evaluate at $x_0 = 1$

• $f(1) = 1^3 - 5(1) + 4 = 1 - 5 + 4 = 0$
• $f'(1) = 3(1)^2 - 5 = 3 - 5 = -2$

Since $f(1) = 0$, the next approximation $x_1$ will be:

$x_1 = 1 - \frac{f(1)}{f'(1)} = 1 - \frac{0}{-2} = 1$

Thus, the tangent line approximation does not change the point in this iteration.

Conclusion:

• The tangent line formula at $x = 1$ is $y = -2(x - 1)$ (since the slope is $f'(1) = -2$).
• The next approximation $x_1 = 1$ remains the same as the starting point $x_0 = 1$ because $f(1) = 0$.

Would you like more detailed steps or have any specific questions?

Here are 5 related questions:

1. What happens in Newton's method if $f(x_0) = 0$?
2. Can you apply Newton's method to a non-differentiable function?
3. How does Newton's method perform if the derivative $f'(x_n)$ is very close to zero?
4. What are the advantages of Newton's method compared to other root-finding algorithms?
5. How many iterations does Newton's method typically take to converge to a root?

Tip: Newton's method requires a good initial guess. If the starting point is too far from the actual root, the method may not converge or may converge to a different root.