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प्रश्न
Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2
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उत्तर
Let x = 2 and Δx = 0.01. Then, we have:
f(2.01) = f(x + Δx) = 4(x + Δx)2 + 5(x + Δx) + 2
Now, Δy = f(x + Δx) − f(x)
∴ f(x + Δx) = f(x) + Δy
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