Finding Taylor Polynomial Approximations of Functions - AP Calculus BC

Let $f(x)=\ln x$ . What is the degree-3 Taylor polynomial centered at $x=1$ ?
Answer: $(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}$ This problem requires finding the degree-3 Taylor polynomial for f(x) = ln x centered at x = 1. Computing derivatives: f(1) = ln(1) = 0, f'(x) = 1/x gives f'(1) = 1, f''(x) = -1/x² gives f''(1) = -1, and f'''(x) = 2/x³ gives f'''(1) = 2. The Taylor polynomial is P₃(x) = 0 + 1·(x-1) + (-1)(x-1)²/2! + 2(x-1)³/3! = (x-1) - (x-1)²/2 + (x-1)³/3. Choice D incorrectly includes a constant term of 1, forgetting that ln(1) = 0. The key insight is that ln x centered at x = 1 has no constant term and follows the pattern (x-1) - (x-1)²/2 + (x-1)³/3 - ..., which is related to ln(1+(x-1)) = ln x.
Find the degree 3 Taylor polynomial for $f(x)=\sqrt{1+x}$ centered at $x=0$ .
Answer: $1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3$ 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.
A damping coefficient is modeled by $f(x)=e^{x}\cos x$ . What is the degree 4 Taylor polynomial about $x=0$ ?
Answer: $1+x-\dfrac{x^3}{3}-\dfrac{x^4}{6}$ 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! $(x-a)^k$ .
For $f(x)=\frac{1}{1-x}$ , what is the degree-4 Taylor polynomial centered at $x=0$ ?
Answer: $1+x+x^2+x^3+x^4$ 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.
Let $f(x)=\arctan x$ . What is the degree-5 Taylor polynomial centered at $x=0$ ?
Answer: $x-\frac{x^3}{3}+\frac{x^5}{5}$ This problem requires finding the degree-5 Taylor polynomial for f(x) = arctan x centered at x = 0. Computing derivatives using f'(x) = 1/(1+x²): f(0) = 0, f'(0) = 1, and higher derivatives follow a pattern. The derivative 1/(1+x²) = 1 - x² + x⁴ - ... leads to f''(0) = 0, f'''(0) = -2, f⁽⁴⁾(0) = 0, and f⁽⁵⁾(0) = 24. Applying the Taylor formula: P₅(x) = 0 + 1·x + 0 + (-2)x³/3! + 0 + 24x⁵/5! = x - x³/3 + x⁵/5. Choice D incorrectly uses -x³/6 instead of -x³/3, confusing the coefficient with sin x. The key pattern is that arctan x has only odd powers with alternating signs and denominators that are the odd numbers themselves: x - x³/3 + x⁵/5 - x⁷/7 + ...
A motion model uses $f(x)=\arctan x$ . What is the degree 7 Taylor polynomial about $x=0$ ?
Answer: $x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}$ Constructing Taylor polynomials involves approximating a function using its derivatives at a specific point, here x=0 for arctan x up to degree 7. The constant term is f(0)=0. The first derivative f'(x)=1/(1+x²), f'(0)=1 for x; higher derivatives lead to odd powers only, with coefficients like -1/3 for x³, 1/5 for x⁵, -1/7 for x⁷. This series has alternating signs for odd terms without factorials in the denominator directly. A tempting distractor like choice E uses sin x factorials, which don't apply to arctan. 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! $(x-a)^k$ .
Let $g(x)=\ln(1+x)$ . What is the degree-5 Taylor polynomial centered at $x=0$ ?
Answer: $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}$ This problem asks for the Taylor polynomial of g(x) = ln(1+x) centered at x = 0. For ln(1+x), we use the fact that successive derivatives follow a pattern: g'(x) = 1/(1+x), g''(x) = -1/ $(1+x)^2$ , g'''(x) = 2/ $(1+x)^3$ , and so on. Evaluating at x = 0 gives g(0) = 0, g'(0) = 1, g''(0) = -1, g'''(0) = 2, $g^{4}$ (0) = -6, $g^{5}$ (0) = 24. The Taylor coefficients are 0, 1/1! = 1, -1/2! = -1/2, 2/3! = 1/3, -6/4! = -1/4, and 24/5! = 1/5. Choice B incorrectly has all positive terms, missing the alternating sign pattern. The general term for ln(1+x) is $(-1)^{n+1}$ $x^n$ /n for n ≥ 1.
For $f(x)=e^{2x}$ , what is the degree-3 Taylor polynomial centered at $x=0$ ?
Answer: $1+2x+2x^2+\frac{4}{3}x^3$ 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.
A population model uses $f(x)=e^{2x}$ . What is the degree 3 Taylor polynomial about $x=0$ ?
Answer: $1+2x+2x^2+\dfrac{4}{3}x^3$ Constructing Taylor polynomials involves approximating a function using its derivatives at a specific point, here $x=0$ for $e^{2x}$ up to degree 3. The constant term is $f(0)=e^0=1$ . The first derivative $f'(x)=2e^{2x}$ , $f'(0)=2$ , giving coefficient $2/1! = 2$ for $x$ ; second $f''(x)=4e^{2x}$ , $f''(0)=4$ , coefficient $4/2! = 2$ for $x^2$ ; third $f'''(x)=8e^{2x}$ , $f'''(0)=8$ , coefficient $8/6=4/3$ for $x^3$ . This reflects the exponential series scaled by powers of 2 in numerators. A tempting distractor like choice B doubles coefficients incorrectly, such as 4 for $x^2$ instead of 2, ignoring the factorial division. To construct Taylor polynomials in general, evaluate the function and its derivatives at the center point and use the formula $P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k$ .
In a statistics model, $f(x)=\ln x$ near $x=1$ . What is the degree 3 Taylor polynomial about $x=1$ ?
Answer: $(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}$ 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 $P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k$ .
For $r(x)=\arctan x$ , what is the degree-7 Taylor polynomial centered at $x=0$ ?
Answer: $x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}$ This problem requires the Taylor polynomial for r(x) = arctan x centered at x = 0. The derivative of arctan x is 1/( $1+x^2$ ), which equals Σ $(-1)^n$ $x^{2n}$ by geometric series. Integrating term by term gives arctan x = Σ $(-1)^n$ $x^{2n+1}$ /(2n+1). Since r(0) = 0 and only odd derivatives are non-zero at x = 0, we get x - $x^3$ /3 + $x^5$ /5 - $x^7$ /7. Choice D incorrectly has - $x^3$ /6 instead of - $x^3$ /3, confusing the coefficient with that of sin x. For arctan x, remember the pattern is odd powers divided by their power number with alternating signs.
Find the degree 4 Taylor polynomial for $f(x)=\ln(1+x)$ centered at $x=0$ .
Answer: $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$ Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For $f(x) = \ln(1+x)$ centered at $x=0$ , the constant term is $f(0) = 0$ , the linear coefficient is $f'(0)/1! = 1$ , the quadratic coefficient is $f''(0)/2! = -1/2$ , the cubic coefficient is $f'''(0)/3! = 2/6 = 1/3$ , and the quartic coefficient is $f''''(0)/4! = -6/24 = -1/4$ . These derivatives are $f'(x) = 1/(1+x)$ , $f''(x) = -1/(1+x)^2$ , $f'''(x) = 2/(1+x)^3$ , and $f''''(x) = -6/(1+x)^4$ , evaluated at $x=0$ . This results in $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$ . A tempting distractor like choice E fails by including an erroneous constant term of 1, which does not match $f(0)=0$ . Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.
Let $f(x)=\frac{1}{1+x}$ . What is the degree $6$ Taylor polynomial centered at $x=0$ ?
Answer: $1-x+x^2-x^3+x^4-x^5+x^6$ This problem asks for the Taylor polynomial of f(x) = 1/(1+x) centered at x = 0. We can rewrite this as $(1+x)^{-1}$ and use the binomial series, or compute derivatives: f(0) = 1, f'(x) = -1/(1+x)² so f'(0) = -1, f''(x) = 2/(1+x)³ so f''(0) = 2, continuing this pattern gives $f^{k}$ (0) = $(-1)^k$ ·k!. Therefore, $P_6$ (x) = 1 - x + 2x²/2! - 6x³/3! + 24x⁴/4! - 120x⁵/5! + 720x⁶/6! = 1 - x + x² - x³ + x⁴ - x⁵ + x⁶. Choice E shows the series for 1/(1+2x) instead, with powers of 2. The key insight is recognizing this as the alternating geometric series 1 - x + x² - x³ + ... for |x| < 1.
Let $f(x)=e^{x}\cos x$ . What is the degree $4$ Taylor polynomial centered at $x=0$ ?
Answer: $1+x-\frac{x^3}{3}-\frac{x^4}{6}$ This problem asks for the Taylor polynomial of f(x) = $e^x$ ·cos(x) centered at x = 0. Using the product rule repeatedly: f(0) = 1·1 = 1, f'(x) = $e^x$ ·cos(x) - $e^x$ ·sin(x) = $e^x$ (cos(x) - sin(x)) so f'(0) = 1, f''(x) = $e^x$ (cos(x) - sin(x)) + $e^x$ (-sin(x) - cos(x)) = - $2e^x$ ·sin(x) so f''(0) = 0, f'''(x) = - $2e^x$ ·sin(x) - $2e^x$ ·cos(x) = - $2e^x$ (sin(x) + cos(x)) so f'''(0) = -2, and $f^{4}$ (x) = - $2e^x$ (sin(x) + cos(x)) - $2e^x$ (cos(x) - sin(x)) = - $4e^x$ ·cos(x) so $f^{4}$ (0) = -4. Therefore, $P_4$ (x) = 1 + x + 0 - 2x³/3! - 4x⁴/4! = 1 + x - x³/3 - x⁴/6. Choice B incorrectly includes -x²/2, missing that f''(0) = 0. The strategy is to carefully apply the product rule and evaluate at x = 0.
A control system uses $f(x)=\dfrac{1}{1+x}$ . What is the degree 5 Taylor polynomial about $x=0$ ?
Answer: $1-x+x^2-x^3+x^4-x^5$ Constructing Taylor polynomials involves approximating a function using its derivatives at a specific point, here $x=0$ for $1/(1+x)$ up to degree 5. The constant term is $f(0)=1$ . The first derivative $f'(x)=-1/(1+x)^2$ , $f'(0)=-1$ for x; second $f''(x)=2/(1+x)^3$ , $f''(0)=2$ , coefficient $2/2=1$ for x $^2$ ; third $=-6/(1+x)^4$ , $-6/6=-1$ for x $^3$ ; alternating thereafter. This is the geometric series with negative ratio. A tempting distractor like choice B uses all positive signs, confusing it with $1/(1-x)$ . To construct Taylor polynomials in general, evaluate the function and its derivatives at the center point and use the formula $P_n(x) = \sum_{k=0}^n f^{(k)}(a) / k! (x-a)^k$ .
For $f(x)=e^{2x}$ , what is the degree-4 Taylor polynomial centered at $x=0$ ?
Answer: $1+2x+2x^2+\frac{4}{3}x^3+\frac{2}{3}x^4$ This problem requires finding the Taylor polynomial for f(x) = $e^{2x}$ centered at x = 0. The Taylor polynomial formula uses $P_n$ (x) = Σ[ $f^{k}$ (0)/k!] $x^k$ , where we need derivatives up to the fourth order. Computing: f(0) = 1, f'(0) = 2, f''(0) = 4, f'''(0) = 8, and $f^{4}$ (0) = 16. The coefficients become 1/0! = 1, 2/1! = 2, 4/2! = 2, 8/3! = 4/3, and 16/4! = 2/3. Choice B incorrectly has 4/3 as the $x^4$ coefficient instead of 2/3, likely from miscomputing 16/24. Remember that for $e^{ax}$ , the nth derivative at x = 0 is $a^n$ , making the coefficient $a^n$ /n!.
Find the degree-4 Taylor polynomial for $f(x)=\cos x$ centered at $x=0$ .
Answer: $1-\frac{x^2}{2}+\frac{x^4}{24}$ This problem requires the degree-4 Taylor polynomial for f(x) = cos x centered at x = 0. The cosine function has only even-powered terms in its Taylor series. Computing derivatives: f(0) = 1, f'(0) = -sin(0) = 0, f''(0) = -cos(0) = -1, f'''(0) = sin(0) = 0, and f⁽⁴⁾(0) = cos(0) = 1. Therefore P₄(x) = 1 + 0 - x²/2 + 0 + x⁴/24 = 1 - x²/2 + x⁴/24. Choice E has x⁴/12 instead of x⁴/24, which is a common factorial error (forgetting that 4! = 24, not 12). The pattern to remember: cos x contains only even powers with alternating signs starting positive.
For $f(x)=\sqrt{1+x}$ , find the degree-3 Taylor polynomial centered at $x=0$ .
Answer: $1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3$ 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.
Let $f(x)=\ln(1+x)$ . What is the degree $5$ Taylor polynomial centered at $x=0$ ?
Answer: $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}$ 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).
Let $p(x)=\cos x$ . Find the degree-6 Taylor polynomial centered at $x=0$ .
Answer: $1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}$ This problem asks for the degree-6 Taylor polynomial of p(x) = cos x centered at x = 0. Computing derivatives: p(0) = 1, p'(x) = -sin x gives p'(0) = 0, p''(x) = -cos x gives p''(0) = -1, p'''(x) = sin x gives p'''(0) = 0, p⁽⁴⁾(x) = cos x gives p⁽⁴⁾(0) = 1, p⁽⁵⁾(x) = -sin x gives p⁽⁵⁾(0) = 0, and p⁽⁶⁾(x) = -cos x gives p⁽⁶⁾(0) = -1. The Taylor polynomial is P₆(x) = 1 + 0 + (-1)x²/2! + 0 + 1·x⁴/4! + 0 + (-1)x⁶/6! = 1 - x²/2 + x⁴/24 - x⁶/720. Choice B incorrectly has all positive terms, missing the alternating pattern of cosine's Taylor series. The key pattern is that cos x contains only even powers with alternating signs: positive for x⁰ and x⁴, negative for x² and x⁶.

Start this quiz