Lagrange Error Bound - AP Calculus BC

1. A Taylor polynomial $P_5$ for $p$ about $0$ estimates $p(0.1)$; if $|p^{(6)}(x)|\le 40$ on $[0,0.1]$, maximum error?
   Answer: $\dfrac{40(0.1)^6}{6!}$ 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 error bound for a degree n polynomial $P_n$(x) at center a is |$R_n$(x)| ≤ (M / (n+1)!) * |x - a|^{n+1}, with M as the max |$p^{(n+1)}$(t)|. For $P_5$ about 0 at 0.1 with |$p^{(6)}$(x)| ≤ 40 on [0,0.1], n=5, M=40, distance 0.1, bound 40*$(0.1)^6$ / 6!. Plug in by verifying n to get the (n+1)th derivative and matching exponent and factorial accordingly. Choice B uses ^5 instead of ^6, failing to account for the full power needed for the remainder term. Always remember to match the factorial to n+1 and the exponent to n+1 for a reliable error estimate in Taylor approximations.
2. For $P_2(x)$ about $a=-1$ with $|f^{(3)}(x)|\le 5$ on $[-1,-0.6]$, what is the maximum error at $x=-0.6$?
   Answer: $\dfrac{5(0.4)^3}{6}$ This question applies the Lagrange error bound to find the maximum approximation error. For a Taylor polynomial Pn(x) centered at a, the error bound 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. Here, P2(x) is centered at a=-1, so n=2, and we evaluate at x=-0.6, giving |x-a|=|-0.6-(-1)|=0.4. With |$f^{3}$(x)| ≤ 5, the maximum error is 5$(0.4)^3$/3! = 5$(0.4)^3$/6. Choice C incorrectly uses 2 in the denominator instead of 3!=6, which happens when students use n instead of (n+1)!. The crucial insight is that the error bound always involves the (n+1)th derivative and (n+1)! for a degree n polynomial.
3. Using $P_4(x)$ about $a=1$ and $|f^{(5)}(x)|\le 40$ on $[1,1.3]$, what is the maximum error at $x=1.3$?
   Answer: $\dfrac{40(0.3)^5}{120}$ 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)!.
4. Using $P_2(x)$ about $a=0.5$ and $|f^{(3)}(x)|\le 30$ on $[0.5,0.9]$, what is the maximum error at $x=0.9$?
   Answer: $\dfrac{30(0.4)^3}{6}$ This question applies the Lagrange error bound to a quadratic Taylor polynomial. For a degree n polynomial Pn(x) centered at a, the error is bounded by |f(x) - Pn(x)| ≤ M|x-a|^(n+1)/(n+1)!. With P2(x) centered at a=0.5, we have n=2, and at x=0.9, we get |x-a|=|0.9-0.5|=0.4. Given |$f^{3}$(x)| ≤ 30, the maximum error is 30$(0.4)^3$/3! = 30$(0.4)^3$/6. Choice D incorrectly uses the fourth power and 4!, which would apply to a cubic (P3) polynomial rather than quadratic (P2). The fundamental rule is that all components in the error bound must correspond to n+1 for a degree n polynomial.
5. Using $P_2$ for $h$ about $x=-2$ to estimate $h(-1.7)$, and $|h^{(3)}(x)|\le 5$ on $[-2,-1.7]$, maximum error?
   Answer: $\dfrac{5(0.3)^3}{3!}$ 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.
6. Using $P_4(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}$ for $e^x$ about $0$, and $|f^{(5)}(x)|\le 3$ on $[0,1]$, what maximum error at $x=1$?
   Answer: $\dfrac{3(1)^5}{5!}$ 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.
7. For $f(x)=e^x$, using $T_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}$, if $|f^{(4)}(x)|\le 3$ on $[0,0.5]$, what is the maximum error at $x=0.5$?
   Answer: $\dfrac{3(0.5)^4}{4!}$ 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.
8. A Taylor polynomial $P_2$ for $t$ about $x=4$ estimates $t(3.8)$; if $|t^{(3)}(x)|\le 30$ on $[3.8,4]$, maximum error?
   Answer: $\dfrac{30(0.2)^3}{3!}$ 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 gives |$R_n$(x)| ≤ (M / (n+1)!) * |x - a|^{n+1} for $P_n$ at a, with M bounding |$t^{(n+1)}$(t)|. For $P_2$ about 4 estimating t(3.8) with |$t^{(3)}$(x)| ≤ 30 on [3.8,4], n=2, M=30, distance 0.2, bound 30*$(0.2)^3$ / 3!. Use by noting quadratic approximation errors with third derivative, power and factorial 3. Choice B uses ^2 instead of ^3, failing to apply the correct exponent for n+1. Always remember to match the factorial to n+1 and the exponent to n+1 for a reliable error estimate in Taylor approximations.
9. For $f(x)=e^{2x}$, using $T_2(x)=1+2x+2x^2$, if $|f^{(3)}(x)|\le 20$ on $[0,0.4]$, what is the maximum error at $x=0.4$?
   Answer: $\dfrac{20(0.4)^3}{3!}$ This problem requires finding the Lagrange error bound for $e^{2x}$ approximated by a degree 2 Taylor polynomial. The error bound uses the 3rd derivative with |$f^{3}$(x)| ≤ 20, since we need the (n+1)th derivative for degree n. At x=0.4, the maximum error is 20$(0.4)^3$/3!, where we use $(x-a)^{n+1}$ = $(0.4-0)^3$ and (n+1)! = 3!. The polynomial $T_2$(x) = 1 + 2x + $2x^2$ correctly captures the first three terms of $e^{2x}$ expanded at x=0. Choice C incorrectly uses 2! instead of 3! in the denominator, confusing the polynomial degree with the factorial needed for the error bound. The consistent rule for Lagrange bounds is that both the exponent and factorial equal (degree + 1).
10. For $P_8(x)$ about $0$ with $|f^{(9)}(x)|\le 3$ on $[-0.2,0.2]$, what is the maximum error at $x=0.2$?
    Answer: $\dfrac{3(0.2)^9}{9!}$ 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).
11. Using $T_5$ centered at $a=-1$ to estimate $f(-0.6)$, with $|f^{(6)}(x)|\le 18$ on $[-1,-0.6]$, find the maximum error.
    Answer: $\dfrac{18(0.4)^6}{6!}$ This problem tests the Lagrange error bound when the Taylor polynomial is centered at a negative value. The error bound formula gives |f(x) - Tₙ(x)| ≤ M|x-a|^(n+1)/(n+1)! for an nth-degree Taylor polynomial. Using T₅ centered at a=-1 to approximate f(-0.6), with |$f^{6}$(x)| ≤ 18, we calculate |error| ≤ 18|-0.6-(-1)|^6/6! = 18|0.4|^6/6! = 18$(0.4)^6$/6!. Notice that |-0.6-(-1)| = |-0.6+1| = |0.4| = 0.4, not 0.6—subtracting the center point correctly is crucial. Always compute |x-a| carefully, especially when dealing with negative numbers.
12. For $\sin x$, using $T_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}$, if $|f^{(6)}(x)|\le 1$ on $[0,0.4]$, what is the maximum error at $x=0.4$?
    Answer: $\dfrac{(0.4)^6}{6!}$ 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.
13. Approximating $f(1.1)$ with $T_0$ about $a=1$, and $|f'(x)|\le 0.8$ on $[1,1.1]$, what is the maximum error?
    Answer: $\dfrac{0.8(0.1)^1}{1!}$ This question tests the Lagrange error bound for the special case of a constant (degree 0) Taylor polynomial. The error bound formula states |f(x) - Tₙ(x)| ≤ M|x-a|^(n+1)/(n+1)! for any degree n. Using T₀ (degree 0, just f(a)) centered at a=1 to approximate f(1.1), with |f'(x)| ≤ 0.8, gives |error| ≤ 0.8|1.1-1|^1/1! = 0.8$(0.1)^1$/1! = 0.8(0.1)/1. Note that 0! would give an undefined expression, and using $(0.1)^0$ = 1 would ignore the distance entirely. For a constant approximation (n=0), the error depends linearly on the distance and uses the first derivative bound.
14. A 2nd-degree Taylor polynomial about $a=0$ estimates $f(0.5)$; if $|f^{(3)}(x)|\le 9$ on $[0,0.5]$, maximum error?
    Answer: $\dfrac{9(0.5)^3}{3!}$ This problem requires applying the Lagrange error bound for a 2nd-degree Taylor polynomial approximation. The Lagrange error bound formula gives $|R_n(x)| \leq \frac{M|x-a|^{n+1}}{(n+1)!}$ where $M$ is the maximum of $|f^{(n+1)}(x)|$ on the interval. For a 2nd-degree polynomial ($n=2$) centered at $a=0$ approximating $f(0.5)$, we get $|R_2(0.5)| \leq \frac{9|0.5-0|^{2+1}}{(2+1)!} = \frac{9(0.5)^3}{3!}$. Choice A incorrectly uses $2!$ in the denominator, which would be appropriate for a 1st-degree polynomial, not a 2nd-degree one. The consistent pattern in Lagrange error bounds is that all three components (derivative order, power, and factorial) are $n+1$.
15. Using $P_3$ about $x=0$ to approximate $s(0.6)$, given $|s^{(4)}(x)|\le 2$ on $[0,0.6]$, what is the maximum error?
    Answer: $\dfrac{2(0.6)^4}{4!}$ 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 $P_n$(x) centered at a, error |$R_n$(x)| ≤ (M / (n+1)!) * |x - a|^{n+1}, M as max |$s^{(n+1)}$(t)|. Using $P_3$ about 0 for s(0.6) with |$s^{(4)}$(x)| ≤ 2 on [0,0.6], n=3, M=2, |x-a|=0.6, bound 2*$(0.6)^4$ / 4!. Apply by ensuring the bound uses the fourth derivative, matching power and factorial to 4. Choice B uses ^3 instead of ^4, which mismatches the n+1 exponent and underestimates. Always remember to match the factorial to n+1 and the exponent to n+1 for a reliable error estimate in Taylor approximations.
16. A Taylor polynomial $P_6(x)$ about $a=-2$ is used, with $|f^{(7)}(x)|\le 210$ on $[-2,-1.5]$; maximum error at $x=-1.5$?
    Answer: $\dfrac{210(0.5)^7}{5040}$ 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)!.
17. Using $P_4(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}$ for $e^x$ to approximate $e^{0.2}$, and $|f^{(5)}(x)|\le 2$ on $[0,0.2]$, what is the maximum error?
    Answer: $\dfrac{2(0.2)^5}{5!}$ 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.
18. For $f(x)=\arctan x$, using $P_3(x)=x-\frac{x^3}{3}$ to approximate $\arctan(0.6)$ and $|f^{(4)}(x)|\le 24$ on $[0,0.6]$, what is the maximum error?
    Answer: $\dfrac{24(0.6)^4}{4!}$ This question involves the Lagrange error bound for an arctangent approximation. The Lagrange error bound formula |Rn(x)| ≤ M|x-a|^(n+1)/(n+1)! connects the polynomial degree n with the (n+1)th derivative bound M. Since P₃(x) has degree 3, we need the fourth derivative (n+1 = 4). Given |$f^{4}$(x)| ≤ 24 on [0,0.6], with center a = 0, the maximum error is 24(0.6)⁴/4!. Choice C mistakenly uses 3! in the denominator, which would be correct for a third derivative but not a fourth. The key principle: the factorial denominator always matches the derivative order, which is one more than the Taylor polynomial's degree.
19. Given $|f^{(5)}(x)|\le 12$ on $[0,0.2]$, what is the maximum error using $T_4$ at $a=0$ for $f(0.2)$?
    Answer: $\dfrac{12(0.2)^5}{5!}$ 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.
20. Using $P_2(x)=x-\frac{x^2}{2}$ for $\ln(1+x)$ to approximate $\ln(0.7)$ and $|f^{(3)}(x)|\le 80$ on $[-0.3,0]$, what is the maximum error?
    Answer: $\dfrac{80(0.3)^3}{3!}$ This problem tests the Lagrange error bound for a logarithm approximation on a negative interval. The Lagrange error bound |Rn(x)| ≤ M|x-a|^(n+1)/(n+1)! applies regardless of whether x < a or x > a. P₂(x) has degree 2, requiring the third derivative bound (n+1 = 3). With |$f^{3}$(x)| ≤ 80 on [-0.3,0] and |x-a| = |-0.3-0| = 0.3, the error bound is 80(0.3)³/3!. Choice C incorrectly uses 2! instead of 3!, mixing up the polynomial degree with the required derivative order. Remember: for a degree n polynomial, always use the (n+1)th derivative and divide by (n+1)! in the error formula.

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