Answer :

Let x = 3 and Δx = 0.02. Then, we get,


f(3.02) = f(x + Δx) = 3(x +Δx)2 + 15( x + Δx) + 5


Now, Δy = f (x + Δx) – f(x)


f (x + Δx) = f(x) + Δy


≈ f(x) + f’(x).Δx (as dx = Δx)


f(3.02) ≈ (3x2 + 15x + 5) + (6x + 15) Δx


= [3(3)2 + 15(3) + 5] +[6(3) + 15] (0.02)


= (27 + 45 + 5) + (18 + 15)(0.02)


= 77 + 0.66


= 77.66


Therefore, the approximate value of f (3.02) is 77.66.

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