a) Which of the following is the correct definition for the
𝑛
th Taylor Polynomial for the function
𝑓
(
𝑥
)
at
𝑥
=
𝑎
? Option III

Option I.
𝑝
𝑛
(
𝑥
)
=
𝑓
(
𝑎
)
+
𝑓
′
(
𝑎
)
2
!
(
𝑥
−
𝑎
)
+
𝑓
″
(
𝑎
)
3
!
(
𝑥
−
𝑎
)
2
+
𝑓
‴
(
𝑎
)
4
!
(
𝑥
−
𝑎
)
3
+
.
.
.
.
+
𝑓
(
𝑛
)
(
𝑎
)
(
𝑛
+
1
)
!
(
𝑥
−
𝑎
)
𝑛

Option II.
𝑝
𝑛
(
𝑥
)
=
𝑓
(
𝑥
)
+
𝑓
′
(
𝑥
)
(
𝑥
−
𝑎
)
+
𝑓
″
(
𝑥
)
2
!
(
𝑥
−
𝑎
)
2
+
𝑓
‴
(
𝑥
)
3
!
(
𝑥
−
𝑎
)
3
+
.
.
.
.
+
𝑓
(
𝑛
)
(
𝑥
)
𝑛
!
(
𝑥
−
𝑎
)
𝑛

Option III.
𝑝
𝑛
(
𝑥
)
=
𝑓
(
𝑎
)
+
𝑓
′
(
𝑎
)
(
𝑥
−
𝑎
)
+
𝑓
″
(
𝑎
)
2
!
(
𝑥
−
𝑎
)
2
+
𝑓
‴
(
𝑎
)
3
!
(
𝑥
−
𝑎
)
3
+
.
.
.
.
+
𝑓
(
𝑛
)
(
𝑎
)
𝑛
!
(
𝑥
−
𝑎
)
𝑛

Option VI.
𝑝
𝑛
(
𝑥
)
=
𝑓
(
𝑎
)
+
𝑓
′
(
𝑎
)
(
𝑥
−
𝑎
)
+
𝑓
″
(
𝑎
)
2
!
(
𝑥
−
𝑎
2
)
+
𝑓
‴
(
𝑎
)
3
!
(
𝑥
−
𝑎
3
)
+
.
.
.
.
+
𝑓
(
𝑛
)
(
𝑎
)
𝑛
!
(
𝑥
−
𝑎
𝑛
)

b) Suppose now that
𝑓
(
𝑥
)
is some differentiable function such that
𝑓
(
2
)
=
10
and that the
𝑛
th derivative of
𝑓
(
𝑥
)
at
𝑥
=
2
is
𝑓
(
𝑛
)
(
2
)
=
2
𝑛
for
1
≤
𝑛
≤
5
. Which of the following would be the 5th Taylor Polynomial of
𝑓
(
𝑥
)
at
𝑥
=
2
? Option 2

Option 1.
𝑝
5
(
𝑥
)
=
10
+
(
𝑥
−
2
)
+
1
2
!
(
𝑥
−
2
)
2
+
1
3
!
(
𝑥
−
2
)
3
+
1
4
!
(
𝑥
−
2
)
4
+
1
5
!
(
𝑥
−
2
)
5

Option 2.
𝑝
5
(
𝑥
)
=
10
+
2
(
𝑥
−
2
)
+
2
2
2
!
(
𝑥
−
2
)
2
+
2
3
3
!
(
𝑥
−
2
)
3
+
2
4
4
!
(
𝑥
−
2
)
4
+
2
5
5
!
(
𝑥
−
2
)
5

Option 3.
𝑝
5
(
𝑥
)
=
2
+
2
(
𝑥
−
2
)
+
2
2
!
(
𝑥
−
2
)
2
+
2
3
!
(
𝑥
−
2
)
3
+
2
4
!
(
𝑥
−
2
)
4
+
2
5
!
(
𝑥
−
2
)
5

Option 4.
𝑝
5
(
𝑥
)
=
10
+
2
(
𝑥
−
2
)
+
2
2
!
(
𝑥
−
2
2
)
2
+
2
3
!
(
𝑥
−
2
3
)
3
+
2
4
!
(
𝑥
−
2
4
)
4
+
2
5
!
(
𝑥
−
2
5
)
5

Question

a) Which of the following is the correct definition for the 
𝑛
th Taylor Polynomial for the function 
𝑓
(
𝑥
)
 at 
𝑥
=
𝑎
 ? Option III

Option I.  
𝑝
𝑛
(
𝑥
)
=
𝑓
(
𝑎
)
+
𝑓
′
(
𝑎
)
2
!
(
𝑥
−
𝑎
)
+
𝑓
″
(
𝑎
)
3
!
(
𝑥
−
𝑎
)
2
+
𝑓
‴
(
𝑎
)
4
!
(
𝑥
−
𝑎
)
3
+
.
.
.
.
+
𝑓
(
𝑛
)
(
𝑎
)
(
𝑛
+
1
)
!
(
𝑥
−
𝑎
)
𝑛

Option II.  
𝑝
𝑛
(
𝑥
)
=
𝑓
(
𝑥
)
+
𝑓
′
(
𝑥
)
(
𝑥
−
𝑎
)
+
𝑓
″
(
𝑥
)
2
!
(
𝑥
−
𝑎
)
2
+
𝑓
‴
(
𝑥
)
3
!
(
𝑥
−
𝑎
)
3
+
.
.
.
.
+
𝑓
(
𝑛
)
(
𝑥
)
𝑛
!
(
𝑥
−
𝑎
)
𝑛

Option III. 
𝑝
𝑛
(
𝑥
)
=
𝑓
(
𝑎
)
+
𝑓
′
(
𝑎
)
(
𝑥
−
𝑎
)
+
𝑓
″
(
𝑎
)
2
!
(
𝑥
−
𝑎
)
2
+
𝑓
‴
(
𝑎
)
3
!
(
𝑥
−
𝑎
)
3
+
.
.
.
.
+
𝑓
(
𝑛
)
(
𝑎
)
𝑛
!
(
𝑥
−
𝑎
)
𝑛

Option VI.  
𝑝
𝑛
(
𝑥
)
=
𝑓
(
𝑎
)
+
𝑓
′
(
𝑎
)
(
𝑥
−
𝑎
)
+
𝑓
″
(
𝑎
)
2
!
(
𝑥
−
𝑎
2
)
+
𝑓
‴
(
𝑎
)
3
!
(
𝑥
−
𝑎
3
)
+
.
.
.
.
+
𝑓
(
𝑛
)
(
𝑎
)
𝑛
!
(
𝑥
−
𝑎
𝑛
)

b) Suppose now that 
𝑓
(
𝑥
)
 is some differentiable function such that 
𝑓
(
2
)
=
10
 and that the 
𝑛
th derivative of  
𝑓
(
𝑥
)
 at 
𝑥
=
2
  is 
𝑓
(
𝑛
)
(
2
)
=
2
𝑛
 for 
1
≤
𝑛
≤
5
 . Which of the following would be the 5th Taylor Polynomial of 
𝑓
(
𝑥
)
 at  
𝑥
=
2
? Option 2

Option 1.  
𝑝
5
(
𝑥
)
=
10
+
(
𝑥
−
2
)
+
1
2
!
(
𝑥
−
2
)
2
+
1
3
!
(
𝑥
−
2
)
3
+
1
4
!
(
𝑥
−
2
)
4
+
1
5
!
(
𝑥
−
2
)
5

Option 2.  
𝑝
5
(
𝑥
)
=
10
+
2
(
𝑥
−
2
)
+
2
2
2
!
(
𝑥
−
2
)
2
+
2
3
3
!
(
𝑥
−
2
)
3
+
2
4
4
!
(
𝑥
−
2
)
4
+
2
5
5
!
(
𝑥
−
2
)
5

Option 3.  
𝑝
5
(
𝑥
)
=
2
+
2
(
𝑥
−
2
)
+
2
2
!
(
𝑥
−
2
)
2
+
2
3
!
(
𝑥
−
2
)
3
+
2
4
!
(
𝑥
−
2
)
4
+
2
5
!
(
𝑥
−
2
)
5

Option 4. 
𝑝
5
(
𝑥
)
=
10
+
2
(
𝑥
−
2
)
+
2
2
!
(
𝑥
−
2
2
)
2
+
2
3
!
(
𝑥
−
2
3
)
3
+
2
4
!
(
𝑥
−
2
4
)
4
+
2
5
!
(
𝑥
−
2
5
)
5

BlackTom AI · Accepted Answer

We start by restating the problem parts to ensure clarity and then evaluate each option step by step.

Part a) Definition of the nth Taylor polynomial for f(x) about x = a.
- Option I presents terms beginning with f(a) and then f'(a) times 2!(x−a), which misplaces the factorial factor for the first derivative term. The general Taylor polynomial uses f^{(k)}(a) / k! times (x−a)^k, so the coefficient for the (x−a) term should be f'(a)/1!, not f'(a)·2!; this makes Option I incorrect.
- Option II writes the polynomial as p_n(x) = f(x) + f'(x)(x−a) + f''(x)^2 (x−a)^2 + ... with derivatives evaluated at x, not at a, and with powers of (x−a) tied to derivatives at x. This misuses the fundamental definition, which requires evaluating derivatives at the expansion point a, not at x, and uses the fixed factorial normalization. Therefore Option II is incorrect.
- Option III shows a pattern that matches the standard Taylor form: f(a) + f'(a)(x−a) + f''(a)/2! (x−a)^2 + f'''(a)/3! (x−a)^3 + ... + f^{(n)}(a)/n! (x−a)^n. Here each term uses the k-th derivative at a divided by k!, multiplied by (x−a)^k, which is exactly the nth Taylor polynomial about a. This makes Option III a correct representation.
- Option VI presents a misformatted or incorrect pattern, with terms like (x−a^n) or (x−a^n) in place of the standard (x−a)^n, which breaks the fundamental definition. Hence Option VI is incorrect.

Part b) Finding the 5th Taylor polynomial p5(x) of f about x = 2 given f(2) = 10 and f^{(n)}(2) = 2n for 1 ≤ n ≤ 5 (and thus f'(2)=2, f''(2)=4, f'''(2)=6, f^{(4)}(2)=8, f^{(5)}(2)=10}). The Taylor polynomial is p5(x) = sum_{k=0}^5 f^{(k)}(2)/k! · (x−2)^k, with f^{(0)}(2) = f(2) = 10.
- The constant term (k=0) is f(2) = 10.
- The linear term (k=1) is f'(2)/1! · (x−2) = 2(x−2).
- The quadratic term (k=2) is f''(2)/2! · (x−2)^2 = 4/2 · (x−2)^2 = 2(x−2)^2.
- The cubic term (k=3) is f'''(2)/3! · (x−2)^3 = 6/6 · (x−2)^3 = 1·(x−2)^3.
- The quartic term (k=4) is f^{(4)}(2)/4! · (x−2)^4 = 8/24 · (x−2)^4 = (1/3)(x−2)^4.
- The quintic term (k=5) is f^{(5)}(2)/5! · (x−2)^5 = 10/120 · (x−2)^5 = (1/12)(x−2)^5.

Now compare the given options with these coefficients:
- Option 1 lists 10 + (x−2) + (1/2)(x−2)^2 + (1/3)(x−2)^3 + (1/4)(x−2)^4 + (1/5)(x−2)^5, which does not match the required coefficients (the linear term should be 2(x−2), and the quadratic Term should be 2(x−2)^2, not 1/2(x−2)^2).
- Option 2 encodes the coefficients as 10 + 2(x−2) + [2 2 2!](x−2)^2 + [2 3 3!](x−2)^3 + [2 4 4!](x−2)^4 + [2 5 5!](x−2)^5, where the middle parts are meant to express f''(2)/2!, f'''(2)/3!, etc. Interpreting the notation as f''(2)/2! = 4/2 = 2, f'''(2)/3! = 6/6 = 1, f^{(4)}(2)/4! = 8/24 = 1/3, f^{(5)}(2)/5! = 10/120 = 1/12, this exactly matches the correct coefficients derived above. Therefore Option 2 is consistent with the correct 5th-degree Taylor polynomial.
- Option 3 starts with a constant term of 2, which already mismatches f(2) = 10, so it cannot be correct.
- Option 4 has misformatted exponents and arguments in the later terms (for example x−2^2, x−2^3, etc.), which do not align with the standard Taylor expansion form and thus are incorrect.

In summary, for part (a) the correct structure corresponds to the standard Taylor form in Option III, and for part (b) the polynomial that correctly reflects the given derivative values through the fifth degree aligns with the interpretation shown in Option 2, when the factorial denominators are applied as in f^{(k)}(2)/k!.

类似问题

Find the Taylor polynomial of order 2 for f(x)=−8x3−x2+x−9 around x=0.

Find the Taylor polynomial of order 2 for 𝑓 ( 𝑥 ) = − 7 𝑥 3 + 9 𝑥 2 − 2 𝑥 − 8 around 𝑥 = 0 .

Let [math: P(x)] be the second-order Taylor polynomial of [math: sin⁡x]\sin {x} at the reference point [math: x=π6]x=\dfrac {\pi }{6}. Compute [math: P(1)]. (Correct the answer to 2 decimal places.)

Let [math: P(x)] be the second-order Taylor polynomial of [math: sin⁡x]\sin {x} at the reference point [math: x=π4]x=\dfrac {\pi }{4}. Compute [math: P(2)]. (Correct the answer to 2 decimal places.)

Let 𝑃 2 ( 𝑥 ) be the quadratic Taylor approximation of the function 𝑓 ( 𝑥 ) = 𝑥 𝑒 2 𝑥 about x=0. What is 𝑃 2 ( 1 ) ?

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