You are using the Newton method to find the minimum of

𝑓
(
𝑥
)
=
3
𝑥
3
+
2
𝑥
−
1

At iteration i, you reached the value

𝑥
𝑖
=
5

What's the value of x at the next iteration?

𝑥
𝑖
+
1
=
?

Question

You are using the Newton method to find the minimum of

𝑓
(
𝑥
)
=
3
𝑥
3
+
2
𝑥
−
1

At iteration i, you reached the value

𝑥
𝑖
=
5

What's the value of x at the next iteration?

𝑥
𝑖
+
1
=
?

BlackTom AI · Accepted Answer

Question restatement: You are using the Newton method to find the minimum of f(x) = 3x^3 + 2x − 1. At iteration i you have x_i = 5. What is x_{i+1}?

Proceeding with Newton's method, the update formula is x_{k+1} = x_k − f'(x_k)/f''(x_k).

First compute the derivatives: f'(x) = 9x^2 + 2 and f''(x) = 18x.

Evaluate at x_i = 5:
- f'(5) = 9*(5)^2 + 2 = 9*25 + 2 = 225 + 2 = 227.
- f''(5) = 18*(5) = 90.

Plug into the update formula:
x_{i+1} = 5 − 227/90.
Compute the fraction: 227/90 ≈ 2.522222...
Subtract from 5: 5 − 2.522222... ≈ 2.477777...

Rounding to three decimal places gives x_{i+1} ≈ 2.478.

Notes and potential pitfalls:
- The Newton update uses the second derivative f''(x) in the denominator; here f''(5) = 90 is nonzero, so the update is valid.
- If you were to present the exact fraction, x_{i+1} = 5 − 227/90 = (450/90 − 227/90) = 223/90 ≈ 2.4777..., which rounds to 2.478 at three decimals.
- Since the provided answer options are missing in the data, this reasoning demonstrates how to obtain the numeric result; if options were listed, one would compare 2.478 to the closest option and select it.

BN5206 Week 4 online quiz

You are using the Newton method to find the minimum of 𝑓 ( 𝑥 ) = 3 𝑥 3 + 2 𝑥 − 1 At iteration i, you reached the value 𝑥 𝑖 = 5 What's the value of x at the next iteration? 𝑥 𝑖 + 1 = ?

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