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Question

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 (dxΔx)
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.

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