Mark M. answered 09/10/17
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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
f(x) = √(x+1) f'(x) = (1/2)(x+1)-½ f"(x) = (-¼)(x+1)-3/2
T1(x) = f(0) + [f'(0)/1!](x-0)1 = 1 + (½)x
T2(x) = f(0) + [f'(0)/1!](x-0)1 + [f"(0)/2!](x-0)2 = 1 + (½)x - (1/8)x2
Mark M.
tutor
You are welcome. If you were asked for a Taylor Polynomial centered at some value other than 0, say 3, you would replace 0 by 3 in the formula.
Mark M (Bayport, NY)
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09/10/17
Sebastian N.
09/10/17