Find the approximate value of f (5.001), where f (x) = x³ – 7x2 + 15.

Let x = 5 and Δx = 0.001. Then, we get,

f(5.001) = f(x + Δx) = (x +Δx)³ - 7 ( x + Δx)² + 15

Now, Δy = f (x + Δx) – f(x)

⇒ f (x + Δx) = f(x) + Δy

≈ f(x) + f’(x).Δx (as dx = Δx)

⇒ f(5.001) ≈ (x³ – 7x² + 15) + (3x² – 14x) Δx

= [(5)² – 7(5)² + 15] +[3(5)² -14(5)] (0.001)

= (125 – 175 + 15) + (75 - 70)(0.001)

= -35 + 0.005

= -34.995

Therefore, the approximate value of f (5.001) is -34.995.