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Find the approximate value of f(2. 01), ...

Find the approximate value of `f(2. 01)`, where `f(x)=4x^2+5x+2`.

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`f(x) = 4x^(2) + 5x +2`
` rArr f'(x) = 8x +5`
Let `x=2 and h=0.01`
Now, ` f(x+h) = f(x) + hf'(x)`
` rArr f(2.01) = f(2) + 0.01 xx f'(2)`
`= [4 xx2^(2) + 5 xx 2 + 2]`
` + 0.01 xx [8 xx 2 + 5]`
` =28 + 0.01 xx 21`
` = 28 + 0.21 = 28.21`
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