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प्रश्न
Find the approximate value of the function f(x) = `sqrt(x^2 + 3x)` at x = 1.02.
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उत्तर
f(x) = `sqrt(x^2 + 3x)`
∴ f'(x) = `d/dx(sqrt(x^2 + 3x))`
= `(1)/(2sqrt(x^2 + 3x)).d/dx(x^2 + 3x)`
= `(1)/(2sqrt(x^2 + 3x)) xx (2x + 3 xx 1)`
= `(2x + 3)/(2sqrt(x^2 + 1)`
Take a = 1 and h = 0.02.
Then f(a) = f(1) = `sqrt(1^2 + 3(1)` = 2
and
f'(a) = f'(1)
= `(2(1) + 3)/(2sqrt(1^2 + 3(1)`
= `(5)/(2 xx 2)`
= `(5)/(4)`
The formula for approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ f(1.02) = f(1 + 0.02)
≑ f(1) + (0.02)f'(1)
≑ `2 + 0.02 xx(5)/(4)`
≑ `(8 + 0.1)/(4)`
= `(8.1)/(4)`
= 2.025
∴ f1.02) ≑ 2.025.
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