Find the approximate value of f(2.002), where f(x) = 2x² + x + 2.

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

Now that Δy = f(x + Δx) - f(x)

Therefore, f(x + Δx) = f(x) + Δy

Calculation: 

Given: f (x) = 2x2 + x + 2.

f'(x) = 4x + 1

Let x + Δx = 2.002 = 2 + 0.002

Therefore, x = 2 and Δx = 0.002

f(x + Δx) = f(x) + Δy

= f(x + Δx) = f(x) + f'(x)Δx

= f(2.002) = 2x2 + x + 2 + (4x + 1)Δx

= f(2.002) = 2(2)² + 2 + 2 + [4⋅(2) + 1](0.002)

= f(2.002) = 12 + (9)(0.002)

= f(2.002) = 12 + 0.018

= f(2.002) = 12.018