AP Calculus BC Quiz: Finding Taylor Polynomial Approximations Of Functions

What this quiz covers

This quiz focuses on Finding Taylor Polynomial Approximations Of Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3$ (correct answer)
2. $1+\frac{1}{2}x+\frac{1}{8}x^2+\frac{1}{16}x^3$
3. $1+x-\frac{1}{2}x^2+\frac{1}{2}x^3$
4. $1+\frac{1}{2}x-\frac{1}{4}x^2+\frac{1}{8}x^3$
5. $1-\frac{1}{2}x-\frac{1}{8}x^2-\frac{1}{16}x^3$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = √(1+x) centered at x=0, the constant term is f(0) = 1, the linear coefficient is f'(0)/1! = 1/2, the quadratic coefficient is f''(0)/2! = -1/4 / 2 = -1/8, and the cubic coefficient is f'''(0)/3! = 3/8 / 6 = 1/16. These derivatives are f'(x) = (1/2)(1+x)^{-1/2}, f''(x) = (1/2)(-1/2)(1+x)^{-3/2} = -1/4 (1+x)^{-3/2}, and f'''(x) = -1/4 (-3/2)(1+x)^{-5/2} = 3/8 (1+x)^{-5/2}, evaluated at x=0. This results in 1 + (1/2)x - (1/8)x^2 + (1/16)x^3. A tempting distractor like choice D fails by using incorrect coefficients, such as -1/4 for quadratic instead of -1/8 after dividing by 2!. Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Question 2

1. $1+x-\dfrac{x^3}{3}-\dfrac{x^4}{6}$ (correct answer)
2. $1+x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{6}$
3. $1+x-\dfrac{x^3}{3}+\dfrac{x^4}{6}$
4. $1+x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{6}$
5. $1+x-\dfrac{x^2}{2}-\dfrac{x^3}{6}-\dfrac{x^4}{24}$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a specific point, here x=0 for e^x cos x up to degree 4. The function value f(0)=1*1=1. Derivatives combine product rule: f'(x)=e^x cos x - e^x sin x, f'(0)=1; f''(0)=0 after calculation; f'''(0)=-2, coefficient -2/6=-1/3 for x³; f^{(4)}(0)=-4, coefficient -4/24=-1/6 for x⁴. This merges exp and cos series, eliminating x² term. A tempting distractor like choice D includes an x² term from cos x alone, ignoring the product cancellation. To construct Taylor polynomials in general, evaluate the function and its derivatives at the center point and use the formula P_n(x) = sum_{k=0}^n f^{(k)}(a) / k! (x-a)^k.

Question 3

1. $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}$ (correct answer)
2. $x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$
3. $x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}$
4. $x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}+\frac{x^5}{120}$
5. $x-\frac{x^2}{3}+\frac{x^3}{4}-\frac{x^4}{5}+\frac{x^5}{6}$

Explanation: This problem asks for the Taylor polynomial of f(x) = ln(1+x) centered at x = 0. Using the Taylor formula, we need derivatives: f(0) = ln(1) = 0, f'(x) = 1/(1+x) so f'(0) = 1, f''(x) = -1/(1+x)² so f''(0) = -1, f'''(x) = 2/(1+x)³ so f'''(0) = 2, f^(4)(x) = -6/(1+x)⁴ so f^(4)(0) = -6, and f^(5)(x) = 24/(1+x)⁵ so f^(5)(0) = 24. Therefore, P_5(x) = 0 + x - x²/2! + 2x³/3! - 6x⁴/4! + 24x⁵/5! = x - x²/2 + x³/3 - x⁴/4 + x⁵/5. Choice D incorrectly divides by factorials in both numerator and denominator, showing confusion about the Taylor formula. The pattern here reveals the alternating series x - x²/2 + x³/3 - x⁴/4 + ..., which matches the known series for ln(1+x).

Question 4

1. $1+x+x^2+x^3+x^4$ (correct answer)
2. $1-x+x^2-x^3+x^4$
3. $1+x+2x^2+3x^3+4x^4$
4. $1+x+x^2+x^3$
5. $1-x-x^2-x^3-x^4$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = 1/(1-x) centered at x=0, the constant term is f(0) = 1, the linear coefficient is f'(0)/1! = 1, the quadratic coefficient is f''(0)/2! = 2/2 = 1, the cubic coefficient is f'''(0)/3! = 6/6 = 1, and the quartic coefficient is f''''(0)/4! = 24/24 = 1. These derivatives increase factorially as f'(x) = 1/(1-x)^2, f''(x) = 2/(1-x)^3, f'''(x) = 6/(1-x)^4, and f''''(x) = 24/(1-x)^5, evaluated at x=0. This yields 1 + x + x^2 + x^3 + x^4. A tempting distractor like choice B fails by alternating signs, which applies to 1/(1+x) instead of 1/(1-x). Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Question 5

1. $1-\frac{x^2}{2}+\frac{x^4}{24}$ (correct answer)
2. $1+\frac{x^2}{2}+\frac{x^4}{24}$
3. $x-\frac{x^3}{6}+\frac{x^5}{120}$
4. $1-\frac{x^2}{2}-\frac{x^4}{24}$
5. $1-\frac{x^2}{4}+\frac{x^4}{48}$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = cos x centered at x=0, the constant term is f(0) = 1, the linear coefficient is f'(0)/1! = 0, the quadratic coefficient is f''(0)/2! = -1/2, the cubic coefficient is f'''(0)/3! = 0, and the quartic coefficient is f''''(0)/4! = 1/24. These derivatives are f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, and f''''(x) = cos x, evaluated at x=0. This results in 1 - (x^2)/2 + (x^4)/24. A tempting distractor like choice D fails by incorrectly making the quartic term negative, ignoring the positive fourth derivative. Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Question 6

1. $(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}$ (correct answer)
2. $1+(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}$
3. $(x-1)+\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}$
4. $(x-1)-\frac{(x-1)^2}{2}-\frac{(x-1)^3}{3}$
5. $(x-1)- (x-1)^2+ (x-1)^3$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = ln x centered at x=1, the constant term is f(1) = 0, the linear coefficient is f'(1)/1! = 1, the quadratic coefficient is f''(1)/2! = -1/2, and the cubic coefficient is f'''(1)/3! = 2/6 = 1/3. These derivatives are f'(x) = 1/x, f''(x) = -1/x^2, and f'''(x) = 2/x^3, evaluated at x=1, with terms in powers of (x-1). This yields (x-1) - (1/2)(x-1)^2 + (1/3)(x-1)^3. A tempting distractor like choice B fails by including an unnecessary constant term of 1, which contradicts f(1)=0. Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Question 7

1. $x-\frac{x^3}{3}+\frac{x^5}{5}$ (correct answer)
2. $x+\frac{x^3}{3}+\frac{x^5}{5}$
3. $x-\frac{x^3}{3}-\frac{x^5}{5}$
4. $1-x+\frac{x^3}{3}-\frac{x^5}{5}$
5. $x-\frac{x^2}{3}+\frac{x^3}{5}$

Explanation: We need to find the fifth-degree Taylor polynomial for f(x) = arctan x centered at x = 0. The derivative f'(x) = 1/(1+x²) can be expanded as a geometric series 1 - x² + x⁴ - x⁶ + ..., and integrating term by term gives arctan x = x - x³/3 + x⁵/5 - x⁷/7 + .... Therefore, the fifth-degree Taylor polynomial is P₅(x) = x - x³/3 + x⁵/5. Choice B incorrectly has all positive signs, missing the alternating pattern that comes from the geometric series expansion of 1/(1+x²). The key strategy is recognizing that arctan x integrates from the geometric series of its derivative.

Question 8

1. $1+x+x^2+x^3+x^4$ (correct answer)
2. $1-x+x^2-x^3+x^4$
3. $1+x+2x^2+3x^3+4x^4$
4. $1+x+x^2+x^3-x^4$
5. $1-x-x^2-x^3-x^4$

Explanation: This problem requires finding the degree-4 Taylor polynomial for f(x) = 1/(1-x) centered at x = 0. We can recognize this as the sum of a geometric series: 1/(1-x) = 1 + x + x² + x³ + x⁴ + ... for |x| < 1. Therefore, the degree-4 Taylor polynomial is simply the first five terms: P₄(x) = 1 + x + x² + x³ + x⁴. We can verify by derivatives: f(0) = 1, f'(x) = 1/(1-x)² gives f'(0) = 1, f''(x) = 2/(1-x)³ gives f''(0) = 2, and so on, yielding coefficients 1, 1, 2/2!, 6/3!, 24/4! = 1, 1, 1, 1, 1. Choice B incorrectly alternates signs, confusing this with 1/(1+x). Remember that 1/(1-x) generates all positive coefficients equal to 1, while 1/(1+x) alternates signs.

Question 9

1. $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$ (correct answer)
2. $x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}$
3. $x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}$
4. $x-\frac{x^2}{3}+\frac{x^3}{3}-\frac{x^4}{4}$
5. $x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}$

Explanation: This problem asks for the degree-4 Taylor polynomial of $g(x) = \ln(1+x)$ centered at $x = 0$. The Taylor polynomial requires computing derivatives at $x = 0$: $g(0) = \ln(1) = 0$, $g'(x) = \frac{1}{1+x}$ gives $g'(0) = 1$, $g''(x) = -\frac{1}{(1+x)^2}$ gives $g''(0) = -1$, $g'''(x) = \frac{2}{(1+x)^3}$ gives $g'''(0) = 2$, and $g^{(4)}(x) = -\frac{6}{(1+x)^4}$ gives $g^{(4)}(0) = -6$. Applying the Taylor formula: $P_4(x) = 0 + 1 \cdot x + (-1) \frac{x^2}{2!} + 2 \frac{x^3}{3!} + (-6) \frac{x^4}{4!} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$. Choice B incorrectly has all positive terms, missing the alternating sign pattern that comes from the derivatives of $\ln(1+x)$. The key insight is that $\ln(1+x)$ generates an alternating series with coefficients $(-1)^{n+1}/n$ for the $x^n$ term.

Question 10

1. $1+2x+2x^2+\frac{4}{3}x^3$ (correct answer)
2. $1+2x+4x^2+\frac{4}{3}x^3$
3. $1+2x+2x^2+\frac{8}{3}x^3$
4. $1+x+2x^2+\frac{4}{3}x^3$
5. $1+2x+x^2+\frac{4}{3}x^3$

Explanation: This problem requires finding the degree-3 Taylor polynomial for f(x) = e^(2x) centered at x = 0. The Taylor polynomial formula uses P₃(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3!. Computing derivatives: f(x) = e^(2x) gives f(0) = 1, f'(x) = 2e^(2x) gives f'(0) = 2, f''(x) = 4e^(2x) gives f''(0) = 4, and f'''(x) = 8e^(2x) gives f'''(0) = 8. Substituting these values yields P₃(x) = 1 + 2x + 4x²/2 + 8x³/6 = 1 + 2x + 2x² + (4/3)x³. Choice B incorrectly uses 4x² instead of 2x², likely forgetting to divide by 2!. Remember that the Taylor polynomial coefficient for x^n is f^(n)(a)/n!, where you must divide the nth derivative by n factorial.

Question 11

1. $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$ (correct answer)
2. $x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$
3. $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$
4. $x-\dfrac{x^3}{3}+\dfrac{x^5}{5}$
5. $x-\dfrac{x^3}{3!}-\dfrac{x^5}{5!}$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a specific point, here x=0 for sin x up to degree 5. The constant term is f(0)=sin(0)=0, so no constant. The first derivative f'(x)=cos x, f'(0)=1, giving coefficient 1 for x; second f''(x)=-sin x, f''(0)=0; third f'''(x)=-cos x, f'''(0)=-1, coefficient -1/3! for x³; fourth f^{(4)}(x)=sin x=0 at 0; fifth f^{(5)}(x)=cos x=1 at 0, coefficient 1/5! for x⁵. This pattern shows odd powers with alternating signs divided by factorials. A tempting distractor like choice D omits factorials, using 1/3 and 1/5 instead of 1/3! and 1/5!, which underestimates the denominator and fails to match the series. To construct Taylor polynomials in general, evaluate the function and its derivatives at the center point and use the formula P_n(x) = sum_{k=0}^n f^{(k)}(a) / k! (x-a)^k.

Question 12

1. $e\left[1+(x-1)+\frac{(x-1)^2}{2}+\frac{(x-1)^3}{6}+\frac{(x-1)^4}{24}\right]$ (correct answer)
2. $e\left[1-(x-1)+\frac{(x-1)^2}{2}-\frac{(x-1)^3}{6}+\frac{(x-1)^4}{24}\right]$
3. $e\left[1+(x-1)+\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\frac{(x-1)^4}{24}\right]$
4. $e\left[1+(x-1)+\frac{(x-1)^2}{4}+\frac{(x-1)^3}{6}+\frac{(x-1)^4}{24}\right]$
5. $\left[1+(x-1)+\frac{(x-1)^2}{2}+\frac{(x-1)^3}{6}+\frac{(x-1)^4}{24}\right]$

Explanation: This problem asks for the degree-4 Taylor polynomial of $f(x) = e^x$ centered at $x = 1$. For exponential functions, all derivatives equal $e^x$, so $f(1) = f'(1) = f''(1) = f'''(1) = f^{(4)}(1) = e$. The Taylor polynomial centered at $x = 1$ uses powers of $(x-1)$: $P_4(x) = e + e(x-1) + \frac{e(x-1)^2}{2!} + \frac{e(x-1)^3}{3!} + \frac{e(x-1)^4}{4!}$. Factoring out e: $P_4(x) = e[1 + (x-1) + \frac{(x-1)^2}{2} + \frac{(x-1)^3}{6} + \frac{(x-1)^4}{24}]$. Choice E incorrectly omits the factor of e, forgetting that all derivatives of $e^x$ evaluated at $x = 1$ equal e, not 1. Remember that when the center is not zero, you must evaluate the function and its derivatives at the center point.

Question 13

1. $1-x+x^2-x^3+x^4$ (correct answer)
2. $1+x+x^2+x^3+x^4$
3. $1-x-x^2-x^3-x^4$
4. $1-2x+3x^2-4x^3+5x^4$
5. $1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = 1/(1+x) centered at x=0, the constant term is f(0) = 1, the linear coefficient is f'(0)/1! = -1, the quadratic coefficient is f''(0)/2! = 2/2 = 1, the cubic coefficient is f'''(0)/3! = -6/6 = -1, and the quartic coefficient is f''''(0)/4! = 24/24 = 1. These derivatives are f'(x) = -1/(1+x)^2, f''(x) = 2/(1+x)^3, f'''(x) = -6/(1+x)^4, and f''''(x) = 24/(1+x)^5, evaluated at x=0, producing alternating signs. This yields 1 - x + x^2 - x^3 + x^4. A tempting distractor like choice B fails by using all positive signs, which applies to 1/(1-x) instead. Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Question 14

1. $x-\frac{x^3}{6}+\frac{x^5}{120}$ (correct answer)
2. $x+\frac{x^3}{6}+\frac{x^5}{120}$
3. $x-\frac{x^3}{6}-\frac{x^5}{120}$
4. $1-\frac{x^2}{2}+\frac{x^4}{24}$
5. $x-\frac{x^3}{3}+\frac{x^5}{60}$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = sin x centered at x=0, the constant term is f(0) = 0, the linear coefficient is f'(0)/1! = 1, the quadratic coefficient is f''(0)/2! = 0, the cubic coefficient is f'''(0)/3! = -1/6, the quartic coefficient is f''''(0)/4! = 0, and the quintic coefficient is f'''''(0)/5! = 1/120. These derivatives cycle as f'(x) = cos x, f''(x) = -sin x, f'''(x) = -cos x, f''''(x) = sin x, and f'''''(x) = cos x, evaluated at x=0. This yields x - (x^3)/6 + (x^5)/120. A tempting distractor like choice C fails by incorrectly making the quintic term negative, which contradicts the positive fifth derivative at zero. Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Question 15

1. $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}$ (correct answer)
2. $1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}$
3. $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$
4. $1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\dfrac{x^6}{120}$
5. $1-\dfrac{x^2}{2!}-\dfrac{x^4}{4!}-\dfrac{x^6}{6!}$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a specific point, here $x=0$ for $\cos x$ up to degree 6. The constant term is $f(0)=\cos(0)=1$. The first derivative $f'(x)=-\sin x$, $f'(0)=0$; second $f''(x)=-\cos x$, $f''(0)=-1$, coefficient $-1/2!$ for $x^2$; third $f'''(0)=0$; fourth $f^{(4)}(0)=1$, coefficient $1/24$ for $x^4$; fifth$=0$; sixth $f^{(6)}(0)=-1$, coefficient $-1/720$ for $x^6$. This yields even powers with alternating signs and factorial denominators. A tempting distractor like choice D omits factorials, using simplified fractions that don't account for the full derivative scaling. To construct Taylor polynomials in general, evaluate the function and its derivatives at the center point and use the formula $P_n(x) = \sum_{k=0}^n f^{(k)}(a) / k! (x-a)^k$.

Question 16

1. $(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}$ (correct answer)
2. $\ln(1)+(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}$
3. $(x-1)+\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}$
4. $(x-1)-\dfrac{(x-1)^2}{2}-\dfrac{(x-1)^3}{3}$
5. $(x-1)-\dfrac{(x-1)^2}{4}+\dfrac{(x-1)^3}{9}$

Explanation: Constructing Taylor polynomials involves approximating a function using its derivatives at a specific point, here $x=1$ for $\ln x$ up to degree 3. The constant term is $f(1) = \ln 1 = 0$. The first derivative $f'(x) = 1/x$, $f'(1) = 1$ for $(x-1)$; second $f''(x) = -1/x^2$, $f''(1) = -1$, coefficient $-1/2$ for $(x-1)^2$; third $f'''(x) = 2/x^3$, $f'''(1) = 2$, coefficient $2/6 = 1/3$ for $(x-1)^3$. Note the center is $x=1$, so terms are in $(x-1)$. A tempting distractor like choice B adds $\ln(1)=0$ unnecessarily as a constant, but it's zero and omitted. To construct Taylor polynomials in general, evaluate the function and its derivatives at the center point and use the formula $P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k$.

Question 17

1. $1+2x+2x^2+\frac{4}{3}x^3$ (correct answer)
2. $1+2x+4x^2+\frac{8}{3}x^3$
3. $1+x+\frac{x^2}{2}+\frac{x^3}{6}$
4. $1+2x+\frac{(2x)^2}{2!}-\frac{(2x)^3}{3!}$
5. $1+2x+2x^2+\frac{8}{3}x^3$

Explanation: This problem asks us to find the third-degree Taylor polynomial for f(x) = e^(2x) centered at x = 0. The Taylor polynomial formula requires us to compute f(0), f'(0), f''(0), and f'''(0), then use P₃(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3!. Since f(x) = e^(2x), we have f'(x) = 2e^(2x), f''(x) = 4e^(2x), and f'''(x) = 8e^(2x), giving us f(0) = 1, f'(0) = 2, f''(0) = 4, and f'''(0) = 8. Substituting into the formula: P₃(x) = 1 + 2x + 4x²/2 + 8x³/6 = 1 + 2x + 2x² + 4x³/3. Choice D incorrectly substitutes 2x into the standard e^x series instead of finding derivatives. The key strategy is to always compute derivatives at the center point before applying the Taylor formula.

Question 18

1. $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}$ (correct answer)
2. $1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}$
3. $1-\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{720}$
4. $x-\frac{x^3}{3!}+\frac{x^5}{5!}$
5. $1-\frac{x^2}{2!}-\frac{x^4}{4!}-\frac{x^6}{6!}$

Explanation: This problem asks for the sixth-degree Taylor polynomial of $f(x) = \cos x$ centered at $x = 0$. The derivatives of $\cos x$ cycle as $\cos x$, $-\sin x$, $-\cos x$, $\sin x$, repeating this pattern, so at $x = 0$ we get: $f(0) = 1$, $f'(0) = 0$, $f''(0) = -1$, $f'''(0) = 0$, $f^{(4)}(0) = 1$, $f^{(5)}(0) = 0$, and $f^{(6)}(0) = -1$. Since odd derivatives are zero, the polynomial contains only even powers: $P_6(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720}$. Choice C incorrectly has a positive $x^6$ term when it should be negative due to the alternating pattern. Remember that $\cos x$ has only even powers with alternating signs in its Taylor series.

Question 19

1. $1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3$ (correct answer)
2. $1+\frac{1}{2}x+\frac{1}{8}x^2+\frac{1}{16}x^3$
3. $1+x-\frac{1}{8}x^2+\frac{1}{16}x^3$
4. $1+\frac{1}{2}x-\frac{1}{4}x^2+\frac{1}{16}x^3$
5. $1+\frac{1}{2}x-\frac{1}{8}x^2-\frac{1}{16}x^3$

Explanation: This problem asks for the degree-3 Taylor polynomial of f(x) = √(1+x) = (1+x)^(1/2) centered at x = 0. Using the binomial series or computing derivatives: f(0) = 1, f'(x) = (1/2)(1+x)^(-1/2) gives f'(0) = 1/2, f''(x) = (-1/4)(1+x)^(-3/2) gives f''(0) = -1/4, and f'''(x) = (3/8)(1+x)^(-5/2) gives f'''(0) = 3/8. The Taylor polynomial is P₃(x) = 1 + (1/2)x + (-1/4)x²/2! + (3/8)x³/3! = 1 + (1/2)x - (1/8)x² + (1/16)x³. Choice B incorrectly has +(1/8)x² instead of -(1/8)x², missing the negative sign from the second derivative. The pattern for (1+x)^(1/2) alternates signs after the first two terms, following the binomial coefficient pattern.

Question 20

1. $1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3$ (correct answer)
2. $1+\frac{1}{2}x+\frac{1}{8}x^2-\frac{1}{16}x^3$
3. $1-\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3$
4. $1+\frac{1}{2}x-\frac{1}{4}x^2+\frac{1}{8}x^3$
5. $1+\frac{1}{2}x-\frac{1}{8}x^2-\frac{1}{16}x^3$

Explanation: This problem asks for the degree-3 Taylor polynomial of f(x) = √(1+x) = (1+x)^(1/2) centered at x = 0. Using the binomial series or computing derivatives: f(0) = 1, f'(x) = (1/2)(1+x)^(-1/2) gives f'(0) = 1/2, f''(x) = (-1/4)(1+x)^(-3/2) gives f''(0) = -1/4, and f'''(x) = (3/8)(1+x)^(-5/2) gives f'''(0) = 3/8. Therefore P₃(x) = 1 + (1/2)x + (-1/4)x²/2 + (3/8)x³/6 = 1 + x/2 - x²/8 + x³/16. Choice D has -x²/4 instead of -x²/8, doubling the second derivative's contribution. Remember: (1+x)^r follows the generalized binomial theorem with coefficients r(r-1)...(r-n+1)/n!.