Find the approximate value of f(2.01), where f(x)=4x2+5x+2.
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Solution
Let x=2 and Δx=0.01. Now, Δy=f(x+Δx)−f(x) f(x+Δx)=f(x)+Δy ≃f(x)+f′(x)⋅Δx, as (dx≃Δx) ⇒f(2.01)≈(4x2+5x+2)+(8x+5)Δx =[4(2)2+5(2)+2]+[8(2)+5](0.01) =28+(21)(0.01)=28+0.21=28.21 Hence, the approximate value of f(2.01) is 28.21.