Show that one of the roots of the equation X3+2x-1=0 lies between x=0 and x=1 and use Newton Raphson method to determine the root, correct to four decimal place.
Answer (2)
Verify the existence of a root between x = 0 and x = 1 by showing that f ( 0 ) and f ( 1 ) have opposite signs, using the Intermediate Value Theorem.
Calculate the derivative of f ( x ) = x 3 + 2 x − 1 , which is f ′ ( x ) = 3 x 2 + 2 .
Apply the Newton-Raphson iterative formula x n + 1 = x n − f ′ ( x n ) f ( x n ) with an initial guess of x 0 = 0.5 until the difference between successive approximations is less than 0.00005.
After a few iterations, the root is found to be approximately 0.4534 .
Explanation
Problem Setup We are given the equation f ( x ) = x 3 + 2 x − 1 = 0 . Our goal is to show that a root exists between x = 0 and x = 1 , and then use the Newton-Raphson method to find the root correct to four decimal places.
Root Existence First, we need to show that a root exists between x = 0 and x = 1 . We can use the Intermediate Value Theorem for this. We evaluate the function at x = 0 and x = 1 .
Applying Intermediate Value Theorem Evaluating f ( 0 ) :
f ( 0 ) = ( 0 ) 3 + 2 ( 0 ) − 1 = − 1 Evaluating f ( 1 ) :
f ( 1 ) = ( 1 ) 3 + 2 ( 1 ) − 1 = 1 + 2 − 1 = 2 Since f ( 0 ) = − 1 and f ( 1 ) = 2 , the function changes sign between x = 0 and x = 1 . Therefore, by the Intermediate Value Theorem, there exists a root between 0 and 1.
Newton-Raphson Method Now, we will use the Newton-Raphson method to find the root. The Newton-Raphson formula is: x n + 1 = x n − f ′ ( x n ) f ( x n ) First, we need to find the derivative of f ( x ) :
f ′ ( x ) = d x d ( x 3 + 2 x − 1 ) = 3 x 2 + 2
Applying the Formula Now we apply the Newton-Raphson iterative formula: x n + 1 = x n − 3 x n 2 + 2 x n 3 + 2 x n − 1 We need to choose an initial guess x 0 between 0 and 1. Let's choose x 0 = 0.5 .
Iterations Iteration 1: x 1 = 0.5 − 3 ( 0.5 ) 2 + 2 ( 0.5 ) 3 + 2 ( 0.5 ) − 1 = 0.5 − 0.75 + 2 0.125 + 1 − 1 = 0.5 − 2.75 0.125 = 0.5 − 0.0454545... ≈ 0.454545 Iteration 2: x 2 = 0.454545 − 3 ( 0.454545 ) 2 + 2 ( 0.454545 ) 3 + 2 ( 0.454545 ) − 1 ≈ 0.454545 − 0.6188 + 2 0.09391 + 0.90909 − 1 ≈ 0.454545 − 2.6188 − 0.002995 ≈ 0.454545 + 0.0011436 ≈ 0.453401 Iteration 3: x 3 = 0.453401 − 3 ( 0.453401 ) 2 + 2 ( 0.453401 ) 3 + 2 ( 0.453401 ) − 1 ≈ 0.453401 − 0.6158 + 2 0.09334 + 0.906802 − 1 ≈ 0.453401 − 2.6158 − 0.000 ≈ 0.453398 Since the difference between x 2 and x 3 is less than 0.00005, we can stop here.
Final Answer Therefore, the root of the equation x 3 + 2 x − 1 = 0 correct to four decimal places is approximately 0.4534 .
Examples
Newton-Raphson method is used in many real-world applications, such as finding the optimal design parameters in engineering. For example, when designing a bridge, engineers use numerical methods like Newton-Raphson to find the dimensions that minimize stress and maximize stability. This ensures the bridge is safe and efficient. Similarly, in finance, it can be used to optimize investment portfolios or to find the equilibrium price in a market model.
We showed that there is a root of the equation x 3 + 2 x − 1 = 0 between x = 0 and x = 1 using the Intermediate Value Theorem. Then, using the Newton-Raphson method with an initial guess of 0.5, we found the root to be approximately 0.4534 .
;
