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Solution for Find the Approximate Value of F (5.001), Where F (X) = X3 − 7x2 + 15. - CBSE (Commerce) Class 12 - Mathematics

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Question

Find the approximate value of f (5.001), where f (x) = x3 − 7x2 + 15 ? 

Solution

\[\text { Let }: \]

\[ x = 5\]

\[ x + ∆ x = 5 . 001\]

\[ \Rightarrow ∆ x = 0 . 001\]

\[f\left( x \right) = x^3 - 7 x^2 + 15\]

\[ \Rightarrow y = f\left( x = 3 \right) = 125 - 175 + 15 = - 35\]

\[\text { Now }, y = f\left( x \right)\]

\[ \Rightarrow \frac{dy}{dx} = 3 x^2 - 14x\]

\[ \therefore dy = ∆ y = \frac{dy}{dx}dx = \left( 3 x^2 - 14x \right) \times 0 . 001 = \left( 75 - 70 \right) \times 0 . 001 = 0 . 005\]

\[ \therefore f\left( 5 . 001 \right) = y + ∆ y = - 35 + 0 . 005 = - 34 . 995\]

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Solution Find the Approximate Value of F (5.001), Where F (X) = X3 − 7x2 + 15. Concept: Approximations.
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