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

Solution:

We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable

Let x = 2 and Δx = 0.01

Then,

f (2.01) = f (x + Δx)

Replacing x with x + Δx in the original function.

= 4 ( x + Δx)² + 5( x + Δx) + 2

Δy = f (x + Δx) - f (x)

f (x + Δx) = f (x) + Δy

≈ f (x) + f' (x).Δx

(∵ dx = Δx)

f (2.01) ≈ (4x² + 5x + 2) + (8x - 5) Δx

= [4 (2)² + 5(2) + 2] + [8(2) + 5] (0.01)

[∵ x = 2, Δx = 0.01]

= (16 + 10 + 2) + (16 + 5)(0.01)

= 28 + 21 (0.01)

= 28 + 0.21

= 28.21

Hence, the approximate value of f (2.01) = 28.21

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.4 Question 2

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

Summary:

The approximate value of f (2.01), where f (x) = 4x² + 5x + 2 is 28.21. We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable