This question requires applying the Lagrange error bound with a non-decimal distance value. The error bound states |f(x) - Tₙ(x)| ≤ M|x-a|^(n+1)/(n+1)! for an nth-degree Taylor polynomial Tₙ. Using T₄ (degree 4) centered at a=0 to approximate f(π/6), with |f^(5)(x)| ≤ 2, gives |error| ≤ $2|π/6-0|^5$/5! = $2(π/6)^5$/5!. A common error would be using 4! or 6! in the denominator, but the correct factorial is (n+1)! = 5! for a 4th-degree polynomial. When working with special values like π/6, treat them just like any other distance in the error formula.

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 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. In this case, for $e^x$ with P_4(x) about 0, n=4, a=0, x=1, and M=3 for the fifth derivative. Thus, the bound calculates to 3 * $(1)^5$ / 5!, corresponding to choice B. A common mistake like choice A uses 4! instead of 5!, failing because the error for a fourth-degree polynomial relies on the fifth derivative and 5! in the denominator. Always ensure the derivative order and factorial align with n+1 to accurately bound errors in series approximations.

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. Here, for ln x with P_2(x) about 1, n=2, a=1, x=1.2 so |x-a|=0.2, and M=2 for the third derivative. The bound is 2 * $(0.2)^3$ / 3!, as in choice B. Distractor C uses $(0.2)^2$, which is incorrect because the exponent should be n+1=3. When the center is not zero, double-check |x - a| to avoid errors in bound calculations for shifted Taylor series.

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. The formula for the error bound when using a degree n Taylor polynomial P_n(x) centered at a to approximate f(x) is |R_n(x)| ≤ (M / (n+1)!) * |x - $a|^{n+1}$, where M is the maximum value of $|f^{(n+1)}$(t)| on the interval between a and x. In this case, for P_4 about 0 approximating f(0.2) with $|f^{(5)}$(x)| ≤ 12 on [0,0.2], n=4, M=12, and |x-a|=0.2, so the bound is $12*(0.2)^5$ / 5!. To apply it, identify the next derivative after the polynomial degree, find its maximum magnitude, and plug into the formula with the distance raised to the n+1 power. A tempting distractor like choice B uses 4! in the denominator instead of 5!, which would be incorrect because the factorial matches the order of the derivative being bounded. Always remember to match the factorial to n+1 and the exponent to n+1 for a reliable error estimate in Taylor approximations.

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.

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. The formula for the error bound when using a degree n Taylor polynomial P_n(x) centered at a to approximate h(x) is |R_n(x)| ≤ (M / (n+1)!) * |x - $a|^{n+1}$, where M bounds $|h^{(n+1)}$(t)| on the relevant interval. For P_2 about -2 approximating h(-1.7) with $|h^{(3)}$(x)| ≤ 5 on [-2,-1.7], n=2, M=5, |x-a|=0.3, so the bound is $5*(0.3)^3$ / 3!. Apply by noting the interval direction doesn't matter since we use absolute distance, and ensure the derivative order is one higher than n. A distractor like choice A uses 2! instead of 3!, which applies to a lower-degree error and overestimates unnecessarily. 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)!.