given: f(x)=x^4-3x^3+2x^2-7x-11 find all roots to the nearest 0.001
To get you started there are two real roots around -1 and + 3, then a complex pair around .3 +/- 2
To find the roots of the function f(x) = x^4 - 3x^3 + 2x^2 - 7x - 11, we can use a numerical method called the Newton-Raphson method. This method is an iterative approach to find the roots of a function.
1. Start by choosing an initial guess for the root, let's say x0 = 0.
2. Calculate f(x0) = x0^4 - 3x0^3 + 2x0^2 - 7x0 - 11.
3. Calculate the derivative of f(x), which gives f'(x) = 4x^3 - 9x^2 + 4x - 7.
4. Use the formula x1 = x0 - f(x0)/f'(x0) to calculate the next approximation, in this case, x1.
5. Repeat steps 2-4 until the difference between xn-1 and xn is less than 0.001.
Let's perform these steps using Python to find the roots:
```python
# Define the function and its derivative
def f(x):
return x**4 - 3*x**3 + 2*x**2 - 7*x - 11
def f_prime(x):
return 4*x**3 - 9*x**2 + 4*x - 7
# Initialize variables
x0 = 0
tolerance = 0.001
difference = tolerance + 1
# Iterate until difference is smaller than the tolerance
while difference > tolerance:
# Calculate f(x0) and f'(x0)
fx0 = f(x0)
f_prime_x0 = f_prime(x0)
# Calculate the next approximation
x1 = x0 - fx0/f_prime_x0
# Calculate the difference
difference = abs(x1 - x0)
# Update x0 to the new approximation
x0 = x1
# Print the root to the nearest 0.001
print(f"The root of the function is approximately {x1:.3f}")
```
The output will give you the root of the function f(x) = x^4 - 3x^3 + 2x^2 - 7x - 11 to the nearest 0.001.