Mark M. answered • 12/31/23
Tutor
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(953)
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
T3(x) = f(4)/0! (x - 4)0 + f'(4)/1! (x - 4)1 + f"(4)/2! (x - 4)2 + f'''(4)/3! (x - 4)3
Since 0! = 1! = 1, we have:
T3(x) = f(4) + f'(4)(x - 4) + f"(4)/2! (x - 4)2 + f'''(4)/3! (x - 4)3
= 32 + (15/2)(x - 4) + ...
Mark M.
tutor
I evaluated f(4) and f'(4) for you. You still need to evaluate f"(4)/2! and f'''(4)/3! and simplify.
Report
01/01/24
Sam A.
T3(x) = f(4)/0! (x - 4)0 + f'(4)/1! (x - 4)1 + f"(4)/2! (x - 4)2 + f'''(4)/3! (x - 4)3 This doesn't seem to be the solution that the software is looking for... it says that only the variable "x" should be used and fully-evaluated.01/01/24