Question

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

Answer 1

f(x)= 4x^2 + 5x + 2

now put x=2.01 ,we get
f(2.01)= 4(2.01)^2 +5×2.01 + 2
=4×4.0401 + 10.05 + 2
=28.2104