Question

find approximation value of f(2.01), where f(x)=4x²+5x+2

Answer 1

$f(x) = 4 {x}^{2} + 5x + 2 \\ f(2.01) = 4 {(2.01)}^{2} + 5 \times 2.01 + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4 \times 4.0401 + 10.05 + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 16.1604 + 10.05 + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 28.2104 \ = 28$

Answer 2

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