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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

#### 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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#### APPEARS IN

RD Sharma Mathematics for Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) (with solutions)
Chapter 14: Differentials, Errors and Approximations
Q: 11 | Page no. 10

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Solution for question: Find the Approximate Value of F (5.001), Where F (X) = X3 − 7x2 + 15. concept: Approximations. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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