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

f(2.01)≈(4x2+5x+2)+(8x+5)∆x

28+0.21

=28.21

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