Use Newton's method to estimate the solutions of the equation 9x^2 + 3x - 1 = 0. Start with x_0 = -1 for the left solution and x_0 = 1 for the right solution. Find x_2 in each case.
To solve this using Newton's method, we need to apply the following iterative formula:
x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n )
where f ( x ) = 9 x 2 + 3 x − 1 f(x) = 9x^2 + 3x - 1 f ( x ) = 9 x 2 + 3 x − 1
Let's break down the problem.
Define f ( x ) f(x) f ( x ) f ′ ( x ) f'(x) f ′ ( x )
f ( x ) = 9 x 2 + 3 x − 1 f(x) = 9x^2 + 3x - 1 f ( x ) = 9 x 2 + 3 x − 1 f ′ ( x ) = 18 x + 3 f'(x) = 18x + 3 f ′ ( x ) = 18 x + 3
Set up initial guesses:
Left solution: Start with x 0 = − 1 x_0 = -1 x 0 = − 1
Right solution: Start with x 0 = 1 x_0 = 1 x 0 = 1
Compute x 1 x_1 x 1 x 2 x_2 x 2
Let's go through these calculations.
For the Left Solution ( x 0 = − 1 ) (x_0 = -1) ( x 0 = − 1 )
Compute x 1 x_1 x 1
x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} x 1 = x 0 − f ′ ( x 0 ) f ( x 0 ) x 0 = − 1 x_0 = -1 x 0 = − 1
f ( − 1 ) = 9 ( − 1 ) 2 + 3 ( − 1 ) − 1 = 9 − 3 − 1 = 5 f(-1) = 9(-1)^2 + 3(-1) - 1 = 9 - 3 - 1 = 5 f ( − 1 ) = 9 ( − 1 ) 2 + 3 ( − 1 ) − 1 = 9 − 3 − 1 = 5 f ′ ( − 1 ) = 18 ( − 1 ) + 3 = − 18 + 3 = − 15 f'(-1) = 18(-1) + 3 = -18 + 3 = -15 f ′ ( − 1 ) = 18 ( − 1 ) + 3 = − 18 + 3 = − 15
So,
x 1 = − 1 − 5 − 15 = − 1 + 1 3 = − 2 3 x_1 = -1 - \frac{5}{-15} = -1 + \frac{1}{3} = -\frac{2}{3} x 1 = − 1 − − 15 5 = − 1 + 3 1 = − 3 2
Compute x 2 x_2 x 2
Substitute x 1 = − 2 3 x_1 = -\frac{2}{3} x 1 = − 3 2
f ( − 2 3 ) = 9 ( − 2 3 ) 2 + 3 ( − 2 3 ) − 1 = 9 ⋅ 4 9 − 2 − 1 = 4 − 2 − 1 = 1 f\left(-\frac{2}{3}\right) = 9\left(-\frac{2}{3}\right)^2 + 3\left(-\frac{2}{3}\right) - 1 = 9 \cdot \frac{4}{9} - 2 - 1 = 4 - 2 - 1 = 1 f ( − 3 2 ) = 9 ( − 3 2 ) 2 + 3 ( − 3 2 ) − 1 = 9 ⋅ 9 4 − 2 − 1 = 4 − 2 − 1 = 1 f ′ ( − 2 3 ) = 18 ( − 2 3 ) + 3 = − 12 + 3 = − 9 f'\left(-\frac{2}{3}\right) = 18\left(-\frac{2}{3}\right) + 3 = -12 + 3 = -9 f ′ ( − 3 2 ) = 18 ( − 3 2 ) + 3 = − 12 + 3 = − 9
Then,
x 2 = − 2 3 − 1 − 9 = − 2 3 + 1 9 = − 5 9 x_2 = -\frac{2}{3} - \frac{1}{-9} = -\frac{2}{3} + \frac{1}{9} = -\frac{5}{9} x 2 = − 3 2 − − 9 1 = − 3 2 + 9 1 = − 9 5
For the Right Solution ( x 0 = 1 ) (x_0 = 1) ( x 0 = 1 )
Compute x 1 x_1 x 1
Substitute x 0 = 1 x_0 = 1 x 0 = 1
f ( 1 ) = 9 ( 1 ) 2 + 3 ( 1 ) − 1 = 9 + 3 − 1 = 11 f(1) = 9(1)^2 + 3(1) - 1 = 9 + 3 - 1 = 11 f ( 1 ) = 9 ( 1 ) 2 + 3 ( 1 ) − 1 = 9 + 3 − 1 = 11 f ′ ( 1 ) = 18 ( 1 ) + 3 = 18 + 3 = 21 f'(1) = 18(1) + 3 = 18 + 3 = 21 f ′ ( 1 ) = 18 ( 1 ) + 3 = 18 + 3 = 21
So,
x 1 = 1 − 11 21 = 10 21 x_1 = 1 - \frac{11}{21} = \frac{10}{21} x 1 = 1 − 21 11 = 21 10
Compute x 2 x_2 x 2
Substitute x 1 = 10 21 x_1 = \frac{10}{21} x 1 = 21 10
f ( 10 21 ) = 9 ( 10 21 ) 2 + 3 ( 10 21 ) − 1 = 9 ⋅ 100 441 + 30 21 − 1 ≈ 0.13 f\left(\frac{10}{21}\right) = 9\left(\frac{10}{21}\right)^2 + 3\left(\frac{10}{21}\right) - 1 = 9 \cdot \frac{100}{441} + \frac{30}{21} - 1 \approx 0.13 f ( 21 10 ) = 9 ( 21 10 ) 2 + 3 ( 21 10 ) − 1 = 9 ⋅ 441 100 + 21 30 − 1 ≈ 0.13 f ′ ( 10 21 ) = 18 ( 10 21 ) + 3 = 180 21 + 3 = 243 21 ≈ 11.57 f'\left(\frac{10}{21}\right) = 18\left(\frac{10}{21}\right) + 3 = \frac{180}{21} + 3 = \frac{243}{21} \approx 11.57 f ′ ( 21 10 ) = 18 ( 21 10 ) + 3 = 21 180 + 3 = 21 243 ≈ 11.57
Then,
x 2 ≈ 10 21 − 0.13 11.57 ≈ 10 21 − 0.0112 ≈ 0.466 x_2 \approx \frac{10}{21} - \frac{0.13}{11.57} \approx \frac{10}{21} - 0.0112 \approx 0.466 x 2 ≈ 21 10 − 11.57 0.13 ≈ 21 10 − 0.0112 ≈ 0.466
Summary of Results
For the left solution, x 2 ≈ − 5 9 x_2 \approx -\frac{5}{9} x 2 ≈ − 9 5
For the right solution, x 2 ≈ 0.466 x_2 \approx 0.466 x 2 ≈ 0.466
Would you like further clarification or details on any steps?
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Tip: Always check if the derivative f ′ ( x ) f'(x) f ′ ( x )