Q.22 Find the approximate value of f(x)=x³+5x²-2x+3 at x=1.98 .

Answer: f(1.98)≈26.4 Take a=2 , h=-0.02 . f(a+h)≈ f(a)+hf'(a) . f(2)=8+20-4+3=27 . f'(x)=3x²+10x-2 , so f'(2)=12+20-2=30 . f(1.98)≈27+(-0.02)(30)=27-0.6=26.4 .

Chapter: Applications of Derivatives· Mathematics & Statistics Part 2

Find the approximate value of \(f(x)=x^3+5x^2-2x+3\) at \(x=1.98\).

Q.22 Find the approximate value of \(f(x)=x^3+5x^2-2x+3\) at \(x=1.98\).

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

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Answer: \(f(1.98)\approx26.4\)

Take \(a=2\), \(h=-0.02\). \(f(a+h)\approx f(a)+hf'(a)\).

\(f(2)=8+20-4+3=27\). \(f'(x)=3x^2+10x-2\), so \(f'(2)=12+20-2=30\).

\(f(1.98)\approx27+(-0.02)(30)=27-0.6=26.4\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Find the approximate value of \(f(x)=x^3+5x^2-2x+3\) at \(x=1.98\).

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