Posted by Alison on Wednesday, November 30, 2011 at 4:17am.
Taylor's series about a (to x² term):
f(x)=f(a) + f'(a)x/1! + f"(a)x²/2!
For
f(x)=(4/3)(7x+50)=(28x/3)+200/3, and a=2
f'(x)=28/3
f"(x)=0
f(x)=f(2) + f'(2)(x-2)/1! + f"(2)(x-2)²/2!
=(4/3)(64) + 28/3(x-2) + 0(x-2)²/2!
=(4/3)(7x-50)
Try the second one along the same lines, and post if you would like to check the answer, which should be the same as the original function.
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