The Lagrange error bound is a key skill in AP Calculus BC for estimating the maximum error when approximating a function with its Taylor polynomial. The formula for the error |R_n(x)| when using the nth-degree Taylor polynomial centered at a is |R_n(x)| ≤ (M / (n+1)!) |x - $a|^{n+1}$, where M is an upper bound for $|f^{(n+1)}$(ξ)| on the interval between a and x. For cos x with P_2(x) about 0, n=2, a=0, x=0.4, and M=1 for the third derivative. This yields 1 * $(0.4)^3$ / 3!, as in choice C. Distractor E uses $(0.4)^4$, which is incorrect since the exponent must be n+1=3 for a second-degree polynomial. Remember to verify the center a and compute |x - a| accurately for precise error estimation in any Taylor series application.

The Lagrange error bound is a key concept in AP Calculus BC for estimating the maximum error when using a Taylor polynomial to approximate a function. For degree n polynomial P_n(x) centered at a, the bound is |R_n(x)| ≤ (M / (n+1)!) * |x - $a|^{n+1}$, M bounding $|q^{(n+1)}$(t)|. Here, P_1 about 3 for q(2.5) with |q''(x)| ≤ 8 on [2.5,3], n=1, M=8, |x-a|=0.5, bound $8*(0.5)^2$ / 2!. Apply by recognizing linear approximation uses second derivative for error, with power and factorial both 2. Choice C mistakenly uses ^1 instead of ^2, which doesn't match the n+1 requirement. Always remember to match the factorial to n+1 and the exponent to n+1 for a reliable error estimate in Taylor approximations.

This problem requires applying the Lagrange error bound formula to estimate the maximum error when using a Taylor polynomial approximation. The Lagrange error bound states that when approximating f(x) with the nth-degree Taylor polynomial Tₙ(x) centered at a, the error is bounded by |f(x) - Tₙ(x)| ≤ M|x-a|^(n+1)/(n+1)!, where M is the maximum value of |f^(n+1)(c)| on the interval. Here, we're using T₄ (degree 4) to approximate f(0.2) with a=0, so we need the 5th derivative bound: |error| ≤ $12|0.2-0|^5$/5! = $12(0.2)^5$/5!. A common mistake is using 4! in the denominator (matching the degree of the polynomial), but remember that the error bound uses (n+1)! where n is the degree. The key strategy is to identify n (the degree of the Taylor polynomial), then use the (n+1)th derivative bound with (n+1)! in the denominator.

This question requires applying the Lagrange error bound for a second-degree Taylor polynomial approximation. The error bound formula states |f(x) - Tₙ(x)| ≤ M|x-a|^(n+1)/(n+1)!, where M bounds the absolute value of the (n+1)th derivative. Using T₂ (degree 2) centered at a=2 to approximate f(2.3), with |f^(3)(x)| ≤ 5, gives |error| ≤ $5|2.3-2|^3$/3! = $5(0.3)^3$/3!. A common mistake would be using $(2.3)^3$ instead of $(0.3)^3$, forgetting to compute the distance |x-a|. The key insight is that the error depends on how far x is from the center a, not on the value of x itself.

This problem involves finding the maximum error when using a 6th-degree Taylor polynomial to approximate a function value. The Lagrange error bound states that for a Taylor polynomial Tₙ of degree n, the error is bounded by M|x-a|^(n+1)/(n+1)!, where M is the maximum of |f^(n+1)| on the interval. With T₆ centered at a=0 approximating f(0.1), and |f^(7)(x)| ≤ 3, we get |error| ≤ $3|0.1-0|^7$/7! = $3(0.1)^7$/7!. A tempting wrong answer might use 6! in the denominator (matching the degree 6), but the error formula requires (n+1)! = 7!. Remember: for an nth-degree Taylor polynomial, use the (n+1)th derivative bound and divide by (n+1)!.

This problem tests understanding of the Lagrange error bound for approximating sin(x) with a degree 5 Taylor polynomial. The Lagrange error bound states that |R_n(x)| ≤ M|x-a|^(n+1)/(n+1)!, where M bounds the (n+1)th derivative on the interval. For T_5(x) (degree 5), we need the 6th derivative: |f^(6)(x)| ≤ 1 for sin(x). The error at x=0.4 is bounded by $1·(0.4)^6$/6!. Choice A incorrectly uses 5! instead of 6! in the denominator, confusing the polynomial degree with the factorial needed for the error bound. Remember that the Lagrange error bound always uses the factorial of (degree + 1), matching the power of (x-a) in the numerator.

This problem tests understanding of the Lagrange error bound for Taylor polynomials. The error bound formula for a degree n Taylor polynomial centered at a is |f(x) - Pn(x)| ≤ M|x-a|^(n+1)/(n+1)!, where M bounds |f^(n+1)(t)| on the interval. With P4(x) centered at a=1, we have n=4, and we evaluate at x=1.3, giving |x-a|=|1.3-1|=0.3. Since |f^(5)(x)| ≤ 40, the maximum error is $40(0.3)^5$/5! = $40(0.3)^5$/120. Choice C incorrectly uses 24 (which is 4!) instead of 120 (which is 5!), a mistake that occurs when students use n! instead of (n+1)!. Remember that for a degree n polynomial, you need the (n+1)th derivative bound and divide by (n+1)!.

This problem requires applying the Lagrange error bound to estimate the maximum error when approximating $e^x$ with a degree 3 Taylor polynomial. The Lagrange error bound formula is |R_n(x)| ≤ M|x-a|^(n+1)/(n+1)!, where M is the maximum value of |f^(n+1)(x)| on the interval, n is the degree of the polynomial, and a is the center (here a=0). Since we're using T_3(x) (degree 3), we need the 4th derivative bound: |f^(4)(x)| ≤ 3. At x=0.5, the error is bounded by $3(0.5)^4$/4! = $3(0.5)^4$/24. Choice B incorrectly uses n=3 in the denominator instead of (n+1)=4, forgetting that the error involves the next derivative after the polynomial degree. When using Lagrange error bounds, always remember that for a degree n polynomial, you need the (n+1)th derivative and (n+1)! in the denominator.

This problem tests the Lagrange error bound for √(1+x) approximated by a degree 2 Taylor polynomial. The error bound formula requires the maximum of |f^(3)(x)| ≤ 2 on [0,0.2], since we need the (n+1)th derivative for a degree n polynomial. At x=0.2, the maximum error is $2(0.2)^3$/3!, using (x-a)^(n+1) = $(0.2-0)^3$ in the numerator and (n+1)! = 3! in the denominator. Choice A incorrectly uses 2! instead of 3!, forgetting that for a degree 2 polynomial, the error involves the third derivative and 3!. When working with Lagrange bounds, always use (degree + 1) for both the power of (x-a) and the factorial—this ensures the error estimate accounts for the next term beyond your polynomial.

This problem involves the Lagrange error bound for arctan(x) approximated by T_3(x), a degree 3 Taylor polynomial. The Lagrange bound requires the 4th derivative bound |f^(4)(x)| ≤ 24 since we need the (n+1)th derivative. For x=0.3 and center a=0, the error is bounded by $24(0.3)^4$/4! = $24(0.3)^4$/24 = $(0.3)^4$. Notice that the given bound of 24 exactly equals 4!, which simplifies the calculation nicely. Choice C incorrectly uses 3! in the denominator, which would give a larger error bound than necessary—this is a common mistake of using n! instead of (n+1)!. Remember that Lagrange error bounds always involve the factorial of (polynomial degree + 1) to match the derivative order used.