Calculus 2 Quiz: Lagrange Error Bound

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Lagrange Error BoundQuestion 1 of 10

The Taylor series for $f(x) = e^x$ centered at $a = 0$ is used to approximate $e^{0.3}$ using the first four terms of the series. If the Lagrange error bound gives a maximum possible error of $\frac{M}{4!} \cdot (0.3)^4$ where $M$ is the maximum value of $|f^{(4)}(c)|$ on the interval $[0, 0.3]$ , what is the tightest upper bound for the absolute error?

\frac{(0.3)^4 \cdot e^{0.3}}{24}

\frac{(0.3)^4 \cdot e^{0.3}}{4!}

\frac{(0.3)^3 \cdot e^{0.3}}{3!}

\frac{(0.3)^4}{4!}

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Calculus 2 Quiz

Calculus 2 Quiz: Lagrange Error Bound

Practice Lagrange Error Bound in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Lagrange Error Bound, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

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.

All questions

Question 1

$\frac{(0.3)^4 \cdot e^{0.3}}{24}$ (correct answer)
$\frac{(0.3)^4 \cdot e^{0.3}}{4!}$
$\frac{(0.3)^3 \cdot e^{0.3}}{3!}$
$\frac{(0.3)^4}{4!}$

Explanation: For

f(x) = e^x

, we have

f^{(4)}(x) = e^x

. The maximum value of

|f^{(4)}(c)|

[0, 0.3]

occurs at

c = 0.3

, giving

M = e^{0.3}

. The Lagrange error bound is

\frac{M}{4!}|x-a|^4 = \frac{e^{0.3}}{24}(0.3)^4

. Note that

4! = 24

. Choice B uses the same form but doesn't simplify

4! = 24

. Choice C uses the wrong degree term. Choice D omits the maximum value

e^{0.3}

Question 2

A function $f(x)$ has the property that $|f^{(n)}(x)| \leq \frac{3^n}{n}$ for all $x \in [-1, 1]$ and all $n \geq 1$ . If the Taylor polynomial $P_4(x)$ centered at $a = 0$ is used to approximate $f(0.6)$ , what is the Lagrange error bound?

$\frac{3^5 \cdot (0.6)^5}{5 \cdot 5!}$ (correct answer)
$\frac{3^4 \cdot (0.6)^4}{4 \cdot 4!}$
$\frac{3^5 \cdot (0.6)^5}{5!}$
$\frac{3^4 \cdot (0.6)^5}{4 \cdot 5!}$

Explanation: For a degree 4 polynomial, the Lagrange error bound uses the 5th derivative:

\frac{M}{5!}|x-a|^5

where

M

is the maximum of

|f^{(5)}(c)|

[0, 0.6]

. Given the bound

|f^{(5)}(x)| \leq \frac{3^5}{5}

, we have

M = \frac{3^5}{5}

. Therefore, the error bound is

\frac{3^5/(5)}{5!} \cdot (0.6)^5 = \frac{3^5 \cdot (0.6)^5}{5 \cdot 5!}

. Choice B uses the 4th derivative instead of 5th. Choice C omits the factor of 5 in the denominator. Choice D mixes up the derivative order.

Question 3

The function $g(x) = (1+x)^{1/3}$ is approximated by its second-degree Taylor polynomial $P_2(x)$ centered at $a = 0$ to estimate $g(0.1)$ . If $g'''(x) = -\frac{2}{9}(1+x)^{-5/3}$ , what interval should be used to find the maximum of $|g'''(c)|$ for the Lagrange error bound?

$[0, 1.1]$ because we must include the point $1 + x = 1 + 0.1 = 1.1$
$[-0.1, 0.1]$ because the interval must be symmetric about the center $a = 0$
$[0, 0.1]$ because we need the interval from the center to the approximation point (correct answer)
$[0, \infty)$ because the third derivative must be evaluated over its entire domain

Explanation: When working with Taylor polynomial approximations and error bounds, the Lagrange error formula requires you to find the maximum value of the next derivative on the interval between your center point and approximation point. For the Lagrange error bound, you need to evaluate

|g'''(c)|

where

c

lies somewhere between the center

a = 0

and the approximation point

x = 0.1

. This means

c

must be in the interval

[0, 0.1]

. Since

g'''(x) = -\frac{2}{9}(1+x)^{-5/3}

, and this function is decreasing as

x

increases (the base

(1+x)

gets larger, making the negative exponent term smaller in absolute value), the maximum of

|g'''(c)|

occurs at

c = 0

. Answer A is incorrect because including

x = 1.1

extends far beyond the relevant interval. The error bound only requires examining values between your center and approximation point, not points related to

1 + x

. Answer B misunderstands the requirement—the interval doesn't need to be symmetric about the center, just span from center to approximation point. Answer D is wrong because you only need to consider the interval where the error occurs, not the entire domain of the third derivative. The key insight is that Lagrange error bounds always involve the interval between your center and the point you're approximating. Don't get distracted by other values that appear in the problem—focus solely on the span from

a

to your target

x

value.

Question 4

Two students calculate Lagrange error bounds for the same Taylor approximation. Student A finds the bound $\frac{M_1}{n!}|x-a|^n$ while Student B finds $\frac{M_2}{n!}|x-a|^n$ where $M_1 < M_2$ . Both students used correct intervals and derivatives. What is the most likely explanation for the difference?

Student A made an error because all Lagrange error bounds for the same approximation must be identical
Student B made an error in computing the maximum value of the derivative on the required interval
Student A found a tighter (better) bound by using a more precise estimate of the derivative maximum (correct answer)
The difference indicates that one student used a different degree polynomial than the other

Explanation: The Lagrange error bound requires finding the maximum of

|f^{(n+1)}(c)|

on the interval between

a

and

x

. While any upper bound for this maximum will give a valid error bound, finding the actual maximum (or a closer approximation to it) gives a tighter, more useful bound. Student A found a better estimate of the true maximum, resulting in

M_1 < M_2

and thus a smaller error bound. Choice A is incorrect because different valid upper bounds can yield different error bounds. Choice B assumes Student B made an error, but both could be correct with B being less precise. Choice D is ruled out since the problem states they used the same approximation.

Question 5

A student uses the third-degree Taylor polynomial for $h(x) = \ln(1+x)$ centered at $a = 0$ to approximate $\ln(1.2)$ . If the actual error is $0.0008$ and the Lagrange error bound predicts a maximum error of $0.0032$ , what can be concluded about the relationship between the actual error and the error bound?

The error bound is invalid because it should equal the actual error exactly in all Taylor approximations
The error bound is working correctly since it provides an upper bound that is greater than the actual error (correct answer)
The error bound is too loose by a factor of 4, indicating an error in the calculation of the fourth derivative
The actual error exceeds the theoretical bound, suggesting the Taylor series does not converge for $x = 0.2$

Explanation: The Lagrange error bound provides an upper bound for the actual error, not an exact value. Since

0.0008 < 0.0032

, the bound is working correctly - it successfully bounds the actual error from above. Choice A is incorrect because error bounds are upper limits, not exact values. Choice C misinterprets the purpose of error bounds. Choice D is wrong because the actual error is less than the bound, and the series does converge for

x = 0.2

Question 6

Consider the Taylor series for $f(x) = \arctan(x)$ centered at $a = 0$ . When using the polynomial $P_5(x) = x - \frac{x^3}{3} + \frac{x^5}{5}$ to approximate $\arctan(0.3)$ , the Lagrange error bound involves $f^{(6)}(c)$ . Given that the derivatives of $\arctan(x)$ become increasingly complex, which approach would give the most practical error bound?

Compute $f^{(6)}(x)$ exactly and find its maximum on $[0, 0.3]$ analytically using calculus techniques
Use the fact that $|f^{(6)}(c)| \leq 6!$ since all derivatives of $\arctan(x)$ are bounded by the corresponding factorial
Evaluate $f^{(6)}(x)$ at several points in $[0, 0.3]$ and use the largest absolute value as an estimate for the maximum
Use the alternating series estimation theorem instead, since $\arctan(x)$ has an alternating series representation (correct answer)

Explanation: For

\arctan(x)

, the Taylor series is alternating with decreasing terms for

|x| < 1

. The alternating series estimation theorem provides a much simpler and often tighter error bound than the Lagrange remainder, especially when the derivatives become complex. Since

0.3 < 1

, this approach is both practical and effective. Choice A is theoretically correct but computationally intensive. Choice B gives an incorrect bound (the derivatives of

\arctan(x)

are not bounded by factorials). Choice C gives only an approximation and may not provide a rigorous upper bound.

Question 7

Consider the function $g(x) = \cos(x)$ and its Taylor polynomial $P_2(x)$ of degree 2 centered at $a = \frac{\pi}{4}$ . When using $P_2(x)$ to approximate $g\left(\frac{\pi}{4} + 0.1\right)$ , which expression correctly represents the Lagrange error bound?

$\frac{|\cos(c)|}{3!}(0.1)^3$ for some $c \in \left[\frac{\pi}{4}, \frac{\pi}{4} + 0.1\right]$
$\frac{|\sin(c)|}{3!}(0.1)^3$ for some $c \in \left[\frac{\pi}{4}, \frac{\pi}{4} + 0.1\right]$ (correct answer)
$\frac{1}{3!}(0.1)^3$ where the maximum is achieved at $c = \frac{\pi}{4}$
$\frac{|\sin(c)|}{2!}(0.1)^2$ for some $c \in \left[\frac{\pi}{4}, \frac{\pi}{4} + 0.1\right]$

Explanation: For

g(x) = \cos(x)

, we have

g'(x) = -\sin(x)

g''(x) = -\cos(x)

, and

g'''(x) = \sin(x)

. Since we're using a degree 2 polynomial, the error involves the third derivative:

g'''(c) = \sin(c)

. The Lagrange error bound is

\frac{|g'''(c)|}{3!}|x-a|^3 = \frac{|\sin(c)|}{3!}(0.1)^3

. Choice A uses

\cos(c)

instead of

\sin(c)

. Choice C assumes the maximum is 1, which isn't necessarily true on this interval. Choice D uses the wrong degree and derivative.

Question 8

The Taylor polynomial $P_n(x)$ of degree $n$ for $f(x) = e^{-x^2}$ centered at $a = 0$ is used to approximate $f(0.5)$ . If we want the Lagrange error bound to guarantee that the approximation error is less than $10^{-6}$ , and we know that $|f^{(n+1)}(c)| \leq 2$ for all $c \in [0, 0.5]$ , what is the minimum degree $n$ required?

$n = 5$ because $\frac{2 \cdot (0.5)^6}{6!} < 10^{-6}$ (correct answer)
$n = 6$ because $\frac{2 \cdot (0.5)^7}{7!} < 10^{-6}$
$n = 7$ because $\frac{2 \cdot (0.5)^8}{8!} < 10^{-6}$
$n = 4$ because $\frac{2 \cdot (0.5)^5}{5!} < 10^{-6}$

Explanation: The Lagrange error bound for a degree

n

polynomial is

\frac{M}{(n+1)!}|x-a|^{n+1}

where

M

is the maximum of

|f^{(n+1)}(c)|

. We need

\frac{2 \cdot (0.5)^{n+1}}{(n+1)!} < 10^{-6}

. For

n=5

\frac{2 \cdot (0.5)^6}{6!} = \frac{2 \cdot (1/64)}{720} = \frac{1/32}{720} ≈ 4.3 \times 10^{-5} > 10^{-6}

. Wait, this is still too large. For

n=6

\frac{2 cdot (0.5)^7}{7!} = \frac{2 \cdot (1/128)}{5040} ≈ 3.1 \times 10^{-6} > 10^{-6}

. For

n=7

\frac{2 \cdot (0.5)^8}{8!} = \frac{2 \cdot (1/256)}{40320} ≈ 1.9 \times 10^{-7} < 10^{-6}

. Therefore

n=7

is the minimum required degree.

Question 9

Consider two different Taylor approximations for $\sin(x)$ at $x = 0.2$ : (1) using $P_3(x)$ centered at $a = 0$ , and (2) using $Q_3(x)$ centered at $a = \frac{\pi}{6}$ . Both are degree 3 polynomials. Which statement about their respective Lagrange error bounds is correct?

Both error bounds are identical since they use the same degree polynomial and the same function
The error bound for $P_3(x)$ is smaller because $|0.2 - 0| < |0.2 - \frac{\pi}{6}|$ (correct answer)
The error bound for $Q_3(x)$ is smaller because $|0.2 - \frac{\pi}{6}| < |0.2 - 0|$
The error bounds depend only on the fourth derivative values and are independent of the distance $|x - a|$

Explanation: The Lagrange error bound is

\frac{M}{(n+1)!}|x-a|^{n+1}

where

M

is the maximum of

|f^{(4)}(c)|

on the relevant interval. Since

0.2 ≈ 0.2

and

\frac{\pi}{6} ≈ 0.524

, we have

|0.2 - 0| = 0.2 < |0.2 - \frac{\pi}{6}| ≈ 0.324

. The term

|x-a|^4

will be smaller for the polynomial centered at 0, making that error bound smaller (assuming similar derivative bounds). Choice A ignores the

|x-a|^{n+1}

term. Choice C has the inequality backwards. Choice D incorrectly states the error bound doesn't depend on

|x-a|

Question 10

For the function $f(x) = \frac{1}{1-x}$ , a fifth-degree Taylor polynomial $P_5(x)$ centered at $a = 0$ is used to approximate $f(0.4)$ . To find the Lagrange error bound, we need the maximum value of $|f^{(6)}(c)|$ on $[0, 0.4]$ . Given that $f^{(6)}(x) = \frac{6!}{(1-x)^7}$ , where does this maximum occur and what is its value?

Maximum occurs at $c = 0$ with value $6!$
Maximum occurs at $c = 0.4$ with value $\frac{6!}{(1.4)^7}$
Maximum occurs at $c = 0.2$ with value $\frac{6!}{(0.8)^7}$
Maximum occurs at $c = 0.4$ with value $\frac{6!}{(0.6)^7}$ (correct answer)

Explanation: When finding Lagrange error bounds for Taylor polynomials, you need to determine where the next derivative (beyond your polynomial's degree) reaches its maximum absolute value on the given interval. This maximum directly affects how large your approximation error could be. For

f^{(6)}(x) = \frac{6!}{(1-x)^7}

, you need to find where

|f^{(6)}(x)|

is maximized on

[0, 0.4]

. Since

6! > 0

and

(1-x)^7 > 0

for all

x

in this interval,

f^{(6)}(x)

is always positive, so

|f^{(6)}(x)| = f^{(6)}(x)

. To maximize

\frac{6!}{(1-x)^7}

, you need to minimize the denominator

(1-x)^7

. As

x

increases from 0 to 0.4, the term

(1-x)

decreases from 1 to 0.6, making

(1-x)^7

smaller. Therefore, the maximum occurs at the right endpoint

x = 0.4

, giving

\frac{6!}{(1-0.4)^7} = \frac{6!}{(0.6)^7}

. Looking at the wrong answers: A incorrectly assumes the maximum occurs at

x = 0

, which actually gives the minimum value since the denominator is largest there. B has the correct location but miscalculates the denominator as

(1.4)^7

instead of

(0.6)^7

. C incorrectly places the maximum at the midpoint and uses the wrong denominator value. Strategy tip: For rational functions with positive numerators, remember that maximizing the function means minimizing the denominator. Always check the behavior at interval endpoints, especially when the function is monotonic.