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Find the approximate values of : f(x) = x3 + 5x2 – 7x + 10 at x = 1.12.

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प्रश्न

Find the approximate values of : f(x) = x3 + 5x2 – 7x + 10 at x = 1.12.

बेरीज
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उत्तर

f(x) = x3 + 5x2 – 7x + 10

∴ f'(x) = `d/dx(x^3 + 5x^2 - 7x + 10)`

= 3x2 + 5 x 2x – 7 x 1 + 0
= 3x2 + 10x – 7
Take a = 1, h = 0.12
Then f(a) = f(1)
= (1)3 + 5(1)2 – 7(1) + 10
= 1 + 5 – 7 + 10
= 9
and
f'(a) = f'(1)
= 3(1)2 + 10(1) – 7
= 3 + 10 – 7
= 6
The formula for approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ f(1.12) = f(1 + 0.12)
≑ f(1) + (0.12).f'(1)
≑ 9 + 0.12 x 6
≑ 9 + 0.72
= 9.72
∴ f(1.12) ≑ 9.72.

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पाठ 2: Applications of Derivatives - Exercise 2.2 [पृष्ठ ७५]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 2 Applications of Derivatives
Exercise 2.2 | Q 6.2 | पृष्ठ ७५

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