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AP Calculus BC Quiz
Practice Lagrange Error Bound in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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For f(x)=cosx, use P2(x)=1−2x2 about 0; if ∣f(3)(x)∣≤1 on [0,0.4], what maximum error at x=0.4?
This quiz focuses on Lagrange Error Bound, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.
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.
For f(x)=cosx, use P2(x)=1−2x2 about 0; if ∣f(3)(x)∣≤1 on [0,0.4], what maximum error at x=0.4?
Explanation: 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.
Using P1 for q about x=3 to approximate q(2.5), and ∣q′′(x)∣≤8 on [2.5,3], maximum error?
Explanation: 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.
Given ∣f(5)(x)∣≤12 on [0,0.2], what is the maximum error using T4 at a=0 for f(0.2)?
Explanation: 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.
Approximating f(2.3) with T2 about a=2, and ∣f(3)(x)∣≤5 on [2,2.3], what is the maximum error?
Explanation: 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.
If ∣f(7)(x)∣≤3 on [−0.1,0.1], what is the maximum error when T6 about 0 estimates f(0.1)?
Explanation: 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)!.
For sinx, using T5(x)=x−3!x3+5!x5, if ∣f(6)(x)∣≤1 on [0,0.4], what is the maximum error at x=0.4?
Explanation: 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.
Using P4(x) about a=1 and ∣f(5)(x)∣≤40 on [1,1.3], what is the maximum error at x=1.3?
Explanation: 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)!.
For f(x)=ex, using T3(x)=1+x+2x2+6x3, if ∣f(4)(x)∣≤3 on [0,0.5], what is the maximum error at x=0.5?
Explanation: 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.
For f(x)=1+x, using T2(x)=1+2x−8x2, if ∣f(3)(x)∣≤2 on [0,0.2], what is the maximum error at x=0.2?
Explanation: 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.
For f(x)=arctanx, using T3(x)=x−3x3, if ∣f(4)(x)∣≤24 on [0,0.3], what is the maximum error at x=0.3?
Explanation: 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.
For P8(x) about 0 with ∣f(9)(x)∣≤3 on [−0.2,0.2], what is the maximum error at x=0.2?
Explanation: This question tests the Lagrange error bound for an eighth-degree Taylor polynomial. The error bound formula is |f(x) - Pn(x)| ≤ M|x-a|^(n+1)/(n+1)!, where M is the maximum of |f^(n+1)(t)| on the interval. For P8(x) centered at a=0, we have n=8, and at x=0.2, we get |x-a|=0.2. With |f^(9)(x)| ≤ 3, the maximum error is 3(0.2)^9/9!. Choice C incorrectly uses 8! in the denominator instead of 9!, which occurs when students forget to use (n+1)! for a degree n polynomial. The key principle is that the factorial in the denominator must match the order of the derivative and the power of (x-a).
A Taylor polynomial P6(x) about a=−2 is used, with ∣f(7)(x)∣≤210 on [−2,−1.5]; maximum error at x=−1.5?
Explanation: This question involves the Lagrange error bound for a sixth-degree Taylor polynomial. The error bound formula states |f(x) - Pn(x)| ≤ M|x-a|^(n+1)/(n+1)!, where M bounds |f^(n+1)(t)| on the interval. For P6(x) centered at a=-2, we have n=6, and at x=-1.5, we get |x-a|=|-1.5-(-2)|=0.5. With |f^(7)(x)| ≤ 210, the maximum error is 210(0.5)^7/7! = 210(0.5)^7/5040. Choice D incorrectly uses the 8th power and 8! (40320), which would apply to a P7 polynomial rather than P6. Remember that for a degree n polynomial, all error bound components must correspond to n+1: the (n+1)th derivative, (n+1)th power, and (n+1)!.
Using P2 centered at a=−2 to estimate f(−1.8), if ∣f(3)(x)∣≤15 on [−2,−1.8], maximum error?
Explanation: This problem involves finding the Lagrange error bound for a 2nd-degree Taylor polynomial approximation. The Lagrange error formula states ∣Rn(x)∣≤(n+1)!M∣x−a∣n+1 where n is the polynomial degree and M bounds ∣f(n+1)(x)∣. For P2 centered at a=−2 approximating f(−1.8), we have ∣R2(−1.8)∣≤(2+1)!15∣−1.8−(−2)∣2+1=3!15(0.2)3. Choice A incorrectly uses 2! in the denominator, which would be appropriate for a 1st-degree polynomial rather than 2nd-degree. Always ensure that the factorial matches the derivative order and power, which are both one more than the polynomial degree.
Using P4(x)=1+x+2x2+6x3+24x4 for ex to approximate e0.2, and ∣f(5)(x)∣≤2 on [0,0.2], what is the maximum error?
Explanation: This problem involves the Lagrange error bound for a higher-degree exponential approximation. The Lagrange error bound |Rn(x)| ≤ M|x-a|^(n+1)/(n+1)! requires identifying that P₄(x) has degree 4, so we need the fifth derivative (n+1 = 5). Given |f^(5)(x)| ≤ 2 on [0,0.2], with center a = 0, the error bound is 2(0.2)⁵/5!. Choice B incorrectly keeps 4! in the denominator while using the fifth power, mixing elements from different derivative orders. The consistent pattern: for a degree n Taylor polynomial, use the (n+1)th derivative bound with matching power and factorial.
Approximating f(π/6) with T4 about a=0, given ∣f(5)(x)∣≤2 for x∈[0,π/6], find the maximum error.
Explanation: 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.
Approximate lnx near 1 with P2(x)=(x−1)−2(x−1)2; if ∣f(3)(x)∣≤2 on [1,1.2], what maximum error at x=1.2?
Explanation: 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.
Using P4(x)=1+x+2x2+6x3+24x4 for ex about 0, and ∣f(5)(x)∣≤3 on [0,1], what maximum error at x=1?
Explanation: 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.
For f(x), ∣f(5)(x)∣≤12 on [0,0.2]; using P4 about 0, what maximum error at x=0.2?
Explanation: 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.
Using P2 for h about x=−2 to estimate h(−1.7), and ∣h(3)(x)∣≤5 on [−2,−1.7], maximum error?
Explanation: 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.
A Taylor polynomial P7 for v about 0 estimates v(0.2); if ∣v(8)(x)∣≤100 on [0,0.2], maximum error?
Explanation: 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 bound is |R_n(x)| ≤ (M / (n+1)!) * |x - a|^{n+1} for P_n at a, with M as max |v^{(n+1)}(t)|. For P_7 about 0 estimating v(0.2) with |v^{(8)}(x)| ≤ 100 on [0,0.2], n=7, M=100, distance 0.2, bound 100*(0.2)^8 / 8!. To apply, identify eighth derivative for degree 7, ensuring exponent and factorial are 8. Choice C uses ^7 instead of ^8, which doesn't align with the n+1 power requirement. Always remember to match the factorial to n+1 and the exponent to n+1 for a reliable error estimate in Taylor approximations.