Find the approximate value of f2-01- where fx-4 x2-5 x-2

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 ≈ Hence, the approx ...

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Standard XII

Mathematics

Integration as Antiderivative

Find the appr...

Question

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

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Solution

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

≈

Hence, the approximate value of f (2.01) is 28.21.

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

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 ≈ Hence, the approximate value of f (2.01) is 28.21.

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

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

≈

Hence, the approximate value of f (2.01) is 28.21.