Calculus 2 Quiz: Taylor Polynomial Approximations

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Taylor Polynomial ApproximationsQuestion 1 of 7

If $P_2(x)$ is the second-degree Taylor polynomial for $f(x) = \ln(1 + x)$ centered at $x = 0$ , and $Q_2(x)$ is the second-degree Taylor polynomial for $g(x) = \ln(1 + 2x)$ centered at $x = 0$ , which relationship between $P_2(x)$ and $Q_2(x)$ is correct?

Q_2(x) = 2P_2(x)

for all values of

x

Q_2(x) = P_2(2x)

for all values of

x

Q_2(x) = P_2(x) + P_2(x)

for small values of

x

Q_2(x) = P_2(2x) - \frac{2x^3}{3}

for all values of

x

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Calculus 2 Quiz

Calculus 2 Quiz: Taylor Polynomial Approximations

Practice Taylor Polynomial Approximations 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 Taylor Polynomial Approximations, 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

$Q_2(x) = 2P_2(x)$ for all values of $x$
$Q_2(x) = P_2(2x)$ for all values of $x$ (correct answer)
$Q_2(x) = P_2(x) + P_2(x)$ for small values of $x$
$Q_2(x) = P_2(2x) - \frac{2x^3}{3}$ for all values of $x$

Explanation: For

\ln(1+x)

P_2(x) = x - \frac{x^2}{2}

. For

\ln(1+2x)

: the derivatives at

x=0

are

f'(0) = 2

f''(0) = -4

, so

Q_2(x) = 2x - \frac{4x^2}{2} = 2x - 2x^2

. Note that

P_2(2x) = 2x - \frac{(2x)^2}{2} = 2x - 2x^2 = Q_2(x)

. Choice A gives

2P_2(x) = 2x - x^2 \neq Q_2(x)

. Choice C is the same as A. Choice D adds an incorrect correction term.

Question 2

The third-degree Taylor polynomial for $f(x) = e^{2x}$ centered at $x = 0$ is used to approximate $f(0.1)$ . If the actual value is $e^{0.2} \approx 1.2214$ , what is the absolute error in this approximation?

Approximately 0.0001
Approximately 0.0014 (correct answer)
Approximately 0.0081
Approximately 0.0214

Explanation: The third-degree Taylor polynomial for

e^{2x}

x = 0

P_3(x) = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} = 1 + 2x + 2x^2 + \frac{4x^3}{3}

. At

x = 0.1

P_3(0.1) = 1 + 0.2 + 0.02 + \frac{0.004}{3} \approx 1.2200

. The absolute error is

|1.2214 - 1.2200| \approx 0.0014

. Choice A uses only terms up to

x^2

, choice C incorrectly computes the third derivative term, and choice D subtracts the linear approximation from the actual value.

Question 3

The third-degree Taylor polynomial for $f(x) = \sqrt{1 + x}$ centered at $x = 0$ is used to approximate $\sqrt{1.08}$ . Which calculation setup is correct for finding this approximation?

$P_3(0.08) = 1 + \frac{1}{2}(0.08) - \frac{1}{8}(0.08)^2 + \frac{1}{16}(0.08)^3$ (correct answer)
$P_3(0.08) = 1 + \frac{1}{2}(0.08) - \frac{1}{8}(0.08)^2 - \frac{1}{16}(0.08)^3$
$P_3(0.08) = 1 + (0.08) - \frac{1}{2}(0.08)^2 + \frac{1}{6}(0.08)^3$
$P_3(0.08) = 1 + \frac{1}{2}(0.08) + \frac{1}{8}(0.08)^2 + \frac{1}{16}(0.08)^3$

Explanation: For

f(x) = (1+x)^{1/2}

, we need:

f(0) = 1

f'(x) = \frac{1}{2}(1+x)^{-1/2} \Rightarrow f'(0) = \frac{1}{2}

f''(x) = -\frac{1}{4}(1+x)^{-3/2} \Rightarrow f''(0) = -\frac{1}{4}

f'''(x) = \frac{3}{8}(1+x)^{-5/2} \Rightarrow f'''(0) = \frac{3}{8}

. So

P_3(x) = 1 + \frac{1}{2}x + \frac{-1/4}{2!}x^2 + \frac{3/8}{3!}x^3 = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3

. Choice B has wrong sign on the cubic term. Choice C uses wrong coefficients. Choice D has wrong signs on the quadratic and cubic terms.

Question 4

Two functions $f(x)$ and $g(x)$ have the same second-degree Taylor polynomial $P_2(x) = 2 + 3x - x^2$ centered at $x = 0$ . If $f(0.1) = 2.29$ and the second-degree Taylor polynomial approximation gives $P_2(0.1) = 2.29$ , what can be concluded about $f'''(0)$ ?

$f'''(0) = 0$ because the approximation is exact
$f'''(0)$ cannot be determined from the given information (correct answer)
$f'''(0) \neq 0$ but the third-order error is negligible
$f'''(0) = 0$ because $f(x)$ must be a quadratic polynomial

Explanation: The fact that

f(0.1) = P_2(0.1)

is a coincidence at this particular point and doesn't imply anything about

f'''(0)

. The Taylor remainder theorem shows that

f(0.1) - P_2(0.1) = \frac{f'''(c)}{3!}(0.1)^3

for some

c

between 0 and 0.1. Since this equals zero, we know

f'''(c) = 0

for some

c

, but this doesn't determine

f'''(0)

. Choice A incorrectly assumes the derivative at 0 equals the derivative at the intermediate point. Choice C assumes

f'''(0) \neq 0

without justification. Choice D incorrectly concludes the function must be polynomial.

Question 5

If $P_n(x)$ represents the $n$ th-degree Taylor polynomial for $\sin(x)$ centered at $x = 0$ , which statement about $P_4(x)$ and $P_5(x)$ is correct?

$P_4(x) = P_5(x)$ because $\sin(x)$ has no $x^4$ term in its series
$P_4(x) = P_5(x)$ because $\sin(x)$ has no $x^5$ term in its series
$P_5(x) = P_4(x) + \frac{x^5}{5!}$ with the additional term being positive (correct answer)
$P_5(x) = P_4(x) - \frac{x^5}{5!}$ with the additional term being negative

Explanation: The Taylor series for

\sin(x)

x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

. Since there is no

x^4

term,

P_4(x) = x - \frac{x^3}{6}

. The

x^5

term has coefficient

+\frac{1}{5!}

(positive in the alternating series), so

P_5(x) = x - \frac{x^3}{6} + \frac{x^5}{120} = P_4(x) + \frac{x^5}{5!}

. Choice A incorrectly identifies which term is missing. Choice B is wrong about both the reasoning and which polynomial has the missing term. Choice D has the wrong sign for the

x^5

coefficient.

Question 6

The function $k(x) = \ln(\cos(x))$ is approximated by its second-degree Taylor polynomial centered at $x = 0$ . Which expression represents this polynomial?

$P_2(x) = -\frac{x^2}{2}$ since $\ln(1) = 0$ and odd derivatives vanish (correct answer)
$P_2(x) = 1 - \frac{x^2}{2}$ using the standard logarithm and cosine series
$P_2(x) = x - \frac{x^2}{2}$ from the chain rule applied to known series
$P_2(x) = -\frac{x^2}{2} - \frac{x^4}{12}$ including the next non-zero term

Explanation: We have

k(0) = \ln(\cos(0)) = \ln(1) = 0

. For the derivatives:

k'(x) = \frac{-\sin(x)}{\cos(x)} = -\tan(x)

, so

k'(0) = 0

k''(x) = -\sec^2(x)

, so

k''(0) = -1

. Therefore

P_2(x) = 0 + 0 \cdot x + \frac{-1}{2!}x^2 = -\frac{x^2}{2}

. Choice B incorrectly starts with 1 instead of 0. Choice C uses wrong derivative calculations. Choice D includes a fourth-degree term which exceeds the requested degree, and the coefficient is incorrect anyway.

Question 7

Consider the function $h(x) = e^x \cos(x)$ . If the second-degree Taylor polynomial for $h(x)$ centered at $x = 0$ is $P_2(x) = a_0 + a_1 x + a_2 x^2$ , what is the value of $a_2$ ?

$a_2 = 0$ because the second derivatives of $e^x$ and $\cos(x)$ cancel
$a_2 = 1$ from the product rule applied to the series
$a_2 = -\frac{1}{2}$ from the second derivative evaluation
$a_2 = 0$ from the second derivative evaluation at $x = 0$ (correct answer)

Explanation: Using the product rule:

h'(x) = e^x \cos(x) - e^x \sin(x)

and

h''(x) = e^x \cos(x) - e^x \sin(x) - e^x \sin(x) - e^x \cos(x) = -2e^x \sin(x)

. At

x = 0

h(0) = 1

h'(0) = 1

, and

h''(0) = -2e^0 \sin(0) = 0

. Therefore

a_2 = \frac{h''(0)}{2!} = 0

. Choice A gives the right answer but wrong reasoning. Choice B incorrectly applies the product rule. Choice C assumes

h''(0) = -1

which is incorrect.