Question

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

Answer 1

Let x = 2 and ∆x = 0.01. Then, we have:
f(2.01) = f(x + ∆x) = 4(x + ∆x)² + 5(x + ∆x) + 2
Now, ∆y = f(x + ∆x) − f(x)

∴ f(x + ∆x) = f(x) + ∆y

Hence, the approximate value of f (2.01) is 28.21.