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Suppose #T_4(x) = 7-3(x-2)+7(x-2)^2-6(x-2)^3+8(x-2)^4# is the 4th-degree Taylor polynomial centered at #x=2# for some function f, what is the value of #f^((3))(2)#?

Answer 1

#-36#

#T_4(x)=(f^((0))(2))/(0!)(x-2)^0+(f^((1))(2))/(1!)(x-2)^1+(f^((2))(2))/(2!)(x-2)^2+(f^((3))(2))/(3!)(x-2)^3+(f^((4))(2))/(4!)(x-2)^4#

By comparing the coefficient of the fourth term,

#Rightarrow (f^((3))(2))/(3!)=-6#

By multiplying both sides by #3!#,

#Rightarrow f^((3))(2)=-6cdot3! =-36#

I hope that this was clear.

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Answer 2

William Abdalla

The value of ( f^{(3)}(2) ) can be found by looking at the coefficient of the third-degree term in the Taylor polynomial ( T_4(x) ). In this case, it's the coefficient of ( (x-2)^3 ), which is (-6). So, ( f^{(3)}(2) = -6 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.