Chapter: Numerical Solutions of Transcendental Equations and System of Linear Equations and Expansion of Function.· Applied Mathematics-I

Using the Newton–Raphson method, find the root of the equation \(2x^3-3x+4=0\) lying between \(-2\) and \(-1\), correct to four places of decimals. (asked in MU Dec 2025)

Q.22 Using the Newton–Raphson method, find the root of the equation \(2x^3-3x+4=0\) lying between \(-2\) and \(-1\), correct to four places of decimals. (asked in MU Dec 2025)

Doubtly AI Asked 27 Oct 2026 AI-generated · not yet verified Practice test mu engineering Save WhatsApp Report

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Answer: \(x\approx-1.6474\).

\(f(x)=2x^3-3x+4\), \(f'(x)=6x^2-3\). \(f(-2)=-6<0,\ f(-1)=5>0\), so a root lies in \((-2,-1)\). Take \(x_0=-1.5\).

Formula: \(x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\).

n	\(x_n\)	\(f(x_n)\)	\(f'(x_n)\)	\(x_{n+1}\)
0	−1.5	1.75	10.5	−1.666667
1	−1.666667	−0.259259	13.666667	−1.647696
2	−1.647696	−0.003585	13.289422	−1.647427
3	−1.647427	≈ −0.000001	13.284089	−1.647427

The last two iterates agree to four decimals, so the root is \(x\approx\mathbf{-1.6474}\).

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

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Using the Newton–Raphson method, find the root of the equation \(2x^3-3x+4=0\) lying between \(-2\) and \(-1\), correct to four places of decimals. (asked in MU Dec 2025)

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