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# Solution for Find the Approximate Value of F (2.01), Where F (X) = 4x2 + 5x + 2 ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2 ?

#### Solution

$\text { Let }:$

$x = 2$

$x + ∆ x = 2 . 01$

$\Rightarrow ∆ x = 0 . 01$

$f\left( x \right) = 4 x^2 + 5x + 2$

$\Rightarrow f\left( x = 2 \right) = 16 + 10 + 2 = 28$

$\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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Solution Find the Approximate Value of F (2.01), Where F (X) = 4x2 + 5x + 2 ? Concept: Approximations.
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