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.