The Total solution for NCERT class 6-12
Find the approximate value of f (2.01),where f (x) = 4x2 +5x + 2
Let x = 2 and Δx =0.01. Then, we have:
f(2.01)= f(x + Δx) = 4(x + Δx)2 +5(x + Δx) + 2
Now, Δy = f(x +Δx) − f(x)
∴ f(x + Δx) = f(x)+ Δy
Hence, the approximate value of f (2.01)is 28.21.