AP Calculus BC Practice Quiz: Finding Taylor Polynomial Approximations of Functions

Correct Answer: B

The coefficient of the $n^{th}$ degree term in a Taylor polynomial for a function $f$ centered at $x=a$ is $\frac{f^{(n)}(a)}{n!}$. For the $(x-2)^2$ term, $a=2$ and $n=2$. The coefficient is $\frac{f''(2)}{2!} = \frac{8}{2} = 4$. [cite: 3216]

Correct Answer: B

The coefficient of the $x^3$ term in the Taylor polynomial is $\frac{f'''(0)}{3!}$. We are given that this coefficient is -4. Therefore, $\frac{f'''(0)}{3!} = -4$, which implies $f'''(0) = -4 \cdot 3! = -4 \cdot 6 = -24$. [cite: 3216]

Correct Answer: B

To approximate $f(1.1)$, we substitute $x=1.1$ into the polynomial $P_2(x)$. $P_2(1.1) = 3 + 5(1.1-1) - 2(1.1-1)^2 = 3 + 5(0.1) - 2(0.1)^2 = 3 + 0.5 - 2(0.01) = 3.5 - 0.02 = 3.48$. [cite: 3218]

Correct Answer: B

Taylor polynomials for a function $f$ centered at $x=a$ can be used to approximate function values of $f$ near $x=a$. Since the polynomial is centered at $x=5$, the value closest to 5, which is $x=4.9$, will yield the best approximation. [cite: 3218]

Correct Answer: B

The general form of the second-degree Taylor polynomial centered at $x=0$ is $P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2$. Substituting the given values: $P_2(x) = 4 + (0)x + \frac{-8}{2}x^2 = 4 - 4x^2$. [cite: 3215, 3216]

Correct Answer: D

This is the definition of the coefficient for the $n^{th}$ degree term in a Taylor polynomial for a function $f$ centered at $x=a$. [cite: 3216]

Correct Answer: B

First, construct the polynomial: $P_2(x) = g(3) + g'(3)(x-3) + \frac{g''(3)}{2!}(x-3)^2 = 2 - 1(x-3) + \frac{6}{2}(x-3)^2 = 2 - (x-3) + 3(x-3)^2$. Now, approximate $g(2.9)$ by calculating $P_2(2.9)$: $P_2(2.9) = 2 - (2.9-3) + 3(2.9-3)^2 = 2 - (-0.1) + 3(-0.1)^2 = 2 + 0.1 + 3(0.01) = 2.1 + 0.03 = 2.13$. [cite: 3215, 3218]

Correct Answer: B

In many cases, as the degree of the Taylor polynomial increases, the polynomial becomes a better approximation of the original function over some interval around the center. Therefore, the fourth-degree polynomial generally provides a more accurate approximation than the second-degree one. [cite: 3217]

Correct Answer: B

The coefficient of the $(x-2)$ term is $\frac{f'(2)}{1!} = 1$, so $f'(2)=1$. The coefficient of the $(x-2)^2$ term is $\frac{f''(2)}{2!} = -5$, so $f''(2) = -5 \cdot 2! = -10$. [cite: 3216]

Correct Answer: A

The coefficient of the $x^4$ term is $\frac{h^{(4)}(0)}{4!}$. From the polynomial, this coefficient is $\frac{1}{24}$. So, $\frac{h^{(4)}(0)}{4!} = \frac{1}{24}$. Since $4! = 24$, we have $\frac{h^{(4)}(0)}{24} = \frac{1}{24}$, which implies $h^{(4)}(0) = 1$. [cite: 3216]

Correct Answer: A

A Taylor polynomial is constructed to match the function's value and the values of its derivatives at the center $x=a$. The first term of the polynomial is $f(a)$. When $x=a$ is substituted into the polynomial, all other terms become zero, so $P_n(a) = f(a)$. [cite: 3215]

Correct Answer: A

From the polynomial: The constant term is $f(-1) = 2$. The coefficient of $(x+1)^2$ is $\frac{f''(-1)}{2!}$. Since there is no $(x+1)^2$ term, its coefficient is 0, so $f''(-1) = 0$. The coefficient of $(x+1)^3$ is $\frac{f'''(-1)}{3!} = 5$, so $f'''(-1) = 5 \cdot 3! = 30$. The sum is $f(-1) + f''(-1) + f'''(-1) = 2 + 0 + 30 = 32$. [cite: 3216]

Correct Answer: C

In many cases, as the degree of a Taylor polynomial increases, the $n^{th}$ degree polynomial will approach the original function over some interval. For functions like $\cos(x)$, this interval of convergence is $(-\infty, \infty)$, meaning the approximation gets better over a wider and wider range as the degree $n$ increases. [cite: 3217]

Correct Answer: C

The point $x=a$ around which the Taylor polynomial is constructed is known as the center of the approximation or expansion. [cite: 3215]