To apply the bisection method to the function $f(x) = \sin(x) + x^3 + 1$, we need to find an interval $[a, b]$ where the function changes signs, indicating a root exists. Here's a step-by-step approach:

1. Determine the interval:

   • Choose two values $a$ and $b$ such that $f(a) \cdot f(b) < 0$. For example, we can evaluate $f(x)$ at various points:

   $f(-2) = \sin(-2) + (-2)^3 + 1 \approx -7.909$ $f(-1) = \sin(-1) + (-1)^3 + 1 \approx -1.841$ $f(0) = \sin(0) + 0^3 + 1 = 1$

   The values show $f(-1) < 0$ and $f(0) > 0$, so we can use the interval $[-1, 0]$.

2. Bisection method steps:

   • Calculate the midpoint $c = \frac{a + b}{2}$ and evaluate $f(c)$.
   • Replace $a$ or $b$ with $c$ depending on the sign of $f(c)$.

3. Iteration:

   • Let’s iterate a few times:

   Iteration 1: $c_1 = \frac{-1 + 0}{2} = -0.5$ $f(-0.5) = \sin(-0.5) + (-0.5)^3 + 1 \approx 0.125$ Since $f(-1) < 0$ and $f(-0.5) > 0$, set $b = -0.5$.

   Iteration 2: $c_2 = \frac{-1 + (-0.5)}{2} = -0.75$ $f(-0.75) = \sin(-0.75) + (-0.75)^3 + 1 \approx -0.290$ Since $f(-0.75) < 0$ and $f(-0.5) > 0$, set $a = -0.75$.

   Iteration 3: $c_3 = \frac{-0.75 + (-0.5)}{2} = -0.625$ $f(-0.625) \approx -0.092$ Update $a$ again.

Continue this process until you reach the desired level of accuracy (e.g., $|f(c)| < 0.001$).

Conclusion

You will find a root in the interval $[-1, 0]$ through iterations.

Would you like to see more iterations or need further details on any specific step? Here are some questions to consider:

1. What is the specific root you want to find using the bisection method?
2. Would you like to explore other numerical methods for finding roots, such as Newton's method or the secant method?
3. How does the bisection method compare to other root-finding methods in terms of efficiency?
4. What are the practical applications of finding roots of functions like this?
5. Do you want to visualize the function and its root graphically?

Tip: Always ensure that the function is continuous in the interval you're examining for the bisection method to be valid.