1.

Find the approximate value of f(3.01) when f(x) = 3x2 + 5x + 4.

Answer»

x + ∆x = 3.01, x = 3 

∆x = 0.01 

f(x + ∆x) = f(x) + f'(x) ∆x 

f(3.01) = f(3) + f'(3) (0.01) = 46 + 23 (0.01) = 46.23



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