Question

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

Answer 1

Answer:

Hey mate here refer an attachment and the answer is 28.21

Answer 2

Step-by-step explanation:

f(x+Δx)=f(2.01)f(x+Δx)=f(2.01)

Let f(x)=2f(x)=2

Δx=0.01Δx=0.01

f(2)=4(2)2+5×2+2f(2)=4(2)2+5×2+2

=4×4+5×2+2=4×4+5×2+2

=16+10+2=16+10+2

=28