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प्रश्न
Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2 ?
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उत्तर
\[\text { Let }: \]
\[ x = 2\]
\[x + ∆ x = 2 . 01\]
\[ \Rightarrow ∆ x = 0 . 01\]
\[f\left( x \right) = 4 x^2 + 5x + 2\]
\[ \Rightarrow f\left( x = 2 \right) = 16 + 10 + 2 = 28\]
\[\text { Now,} y = f\left( x \right)\]
\[ \Rightarrow \frac{dy}{dx} = 8x + 5\]
\[ \therefore dy = ∆ y = \frac{dy}{dx}dx = \left( 8x + 5 \right) \times 0 . 01 = \left( 16 + 5 \right) \times 0 . 01 = 0 . 21\]
\[ \therefore f\left( 2 . 01 \right) = y + ∆ y = 28 . 21\]
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