Answer :
To find a real root of the equation [tex]\(x^3 - 6x - 4 = 0\)[/tex] using the Bisection method, we follow a systematic approach. Here's the detailed, step-by-step solution:
### Step 1: Define the function
The function for which we are finding the root is:
[tex]\[ f(x) = x^3 - 6x - 4 \][/tex]
### Step 2: Initial Interval
We need to choose an initial interval [tex]\([a, b]\)[/tex] where the function changes sign (i.e., [tex]\(f(a) \cdot f(b) < 0\)[/tex]). Let's select [tex]\(a = 1\)[/tex] and [tex]\(b = 3\)[/tex].
### Step 3: Check Values at Endpoints
First, evaluate the function at the endpoints of the interval:
[tex]\[ f(1) = 1^3 - 6 \cdot 1 - 4 = -9 \][/tex]
[tex]\[ f(3) = 3^3 - 6 \cdot 3 - 4 = 11 \][/tex]
Since [tex]\(f(1)\)[/tex] is negative and [tex]\(f(3)\)[/tex] is positive ([tex]\(f(1) \cdot f(3) < 0\)[/tex]), there is at least one root in the interval [tex]\([1, 3]\)[/tex].
### Step 4: Calculate Midpoint
Calculate the midpoint of the interval:
[tex]\[ \text{midpoint} = \frac{1 + 3}{2} = 2 \][/tex]
### Step 5: Iterative Bisection Process
To achieve a more accurate root, we repeatedly apply the Bisection method until the interval's length is less than a specified tolerance level, say [tex]\(\text{tolerance} = 10^{-6}\)[/tex]. We follow the iterative process:
1. Evaluate [tex]\(f(\text{midpoint}) = f(2)\)[/tex]:
[tex]\[ f(2) = 2^3 - 6 \cdot 2 - 4 = -8 \][/tex]
2. Determine the new interval based on the sign of [tex]\(f(\text{midpoint})\)[/tex]:
- If [tex]\(f(2) < 0\)[/tex], update [tex]\(a\)[/tex] to the midpoint, and keep [tex]\(b\)[/tex] unchanged, so the new interval becomes [tex]\([2, 3]\)[/tex].
3. Calculate the new midpoint of the next interval:
[tex]\[ \text{midpoint} = \frac{2 + 3}{2} = 2.5 \][/tex]
4. Continue this process, evaluating [tex]\(f(\text{new midpoint})\)[/tex], and refining the interval.
Since the specific tolerance level is [tex]\(10^{-6}\)[/tex], the iteration continues until [tex]\(\frac{b - a}{2}\)[/tex] is sufficiently small and we converge to an approximate root.
### Final Interval and Approximate Root
After several iterations of this process, we narrow down the interval to:
[tex]\[ a \approx 2.7320499420166016 \][/tex]
[tex]\[ b \approx 2.7320518493652344 \][/tex]
And the approximate root (midpoint of this interval) is:
[tex]\[ x \approx 2.732050895690918 \][/tex]
### Conclusion
By using the Bisection method, the real root of the equation [tex]\(x^3 - 6x - 4 = 0\)[/tex] is approximately [tex]\( x \approx 2.7320509 \)[/tex]. This value provides a close approximation to the true root within the specified tolerance.
### Step 1: Define the function
The function for which we are finding the root is:
[tex]\[ f(x) = x^3 - 6x - 4 \][/tex]
### Step 2: Initial Interval
We need to choose an initial interval [tex]\([a, b]\)[/tex] where the function changes sign (i.e., [tex]\(f(a) \cdot f(b) < 0\)[/tex]). Let's select [tex]\(a = 1\)[/tex] and [tex]\(b = 3\)[/tex].
### Step 3: Check Values at Endpoints
First, evaluate the function at the endpoints of the interval:
[tex]\[ f(1) = 1^3 - 6 \cdot 1 - 4 = -9 \][/tex]
[tex]\[ f(3) = 3^3 - 6 \cdot 3 - 4 = 11 \][/tex]
Since [tex]\(f(1)\)[/tex] is negative and [tex]\(f(3)\)[/tex] is positive ([tex]\(f(1) \cdot f(3) < 0\)[/tex]), there is at least one root in the interval [tex]\([1, 3]\)[/tex].
### Step 4: Calculate Midpoint
Calculate the midpoint of the interval:
[tex]\[ \text{midpoint} = \frac{1 + 3}{2} = 2 \][/tex]
### Step 5: Iterative Bisection Process
To achieve a more accurate root, we repeatedly apply the Bisection method until the interval's length is less than a specified tolerance level, say [tex]\(\text{tolerance} = 10^{-6}\)[/tex]. We follow the iterative process:
1. Evaluate [tex]\(f(\text{midpoint}) = f(2)\)[/tex]:
[tex]\[ f(2) = 2^3 - 6 \cdot 2 - 4 = -8 \][/tex]
2. Determine the new interval based on the sign of [tex]\(f(\text{midpoint})\)[/tex]:
- If [tex]\(f(2) < 0\)[/tex], update [tex]\(a\)[/tex] to the midpoint, and keep [tex]\(b\)[/tex] unchanged, so the new interval becomes [tex]\([2, 3]\)[/tex].
3. Calculate the new midpoint of the next interval:
[tex]\[ \text{midpoint} = \frac{2 + 3}{2} = 2.5 \][/tex]
4. Continue this process, evaluating [tex]\(f(\text{new midpoint})\)[/tex], and refining the interval.
Since the specific tolerance level is [tex]\(10^{-6}\)[/tex], the iteration continues until [tex]\(\frac{b - a}{2}\)[/tex] is sufficiently small and we converge to an approximate root.
### Final Interval and Approximate Root
After several iterations of this process, we narrow down the interval to:
[tex]\[ a \approx 2.7320499420166016 \][/tex]
[tex]\[ b \approx 2.7320518493652344 \][/tex]
And the approximate root (midpoint of this interval) is:
[tex]\[ x \approx 2.732050895690918 \][/tex]
### Conclusion
By using the Bisection method, the real root of the equation [tex]\(x^3 - 6x - 4 = 0\)[/tex] is approximately [tex]\( x \approx 2.7320509 \)[/tex]. This value provides a close approximation to the true root within the specified tolerance.
