AP Calculus BC Flashcards: Local Linearity And Linearization

What this deck covers

This deck focuses on Local Linearity And Linearization, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the linear approximation of $f(x) = \frac{1}{x}$ at $x = 3$?

Answer: $L(x) = \frac{1}{3} - \frac{1}{9}(x-3)$. $f(3) = \frac{1}{3}$, $f'(3) = -\frac{1}{9}$, standard linearization formula.

Flashcard 2: Find the linear approximation of $f(x) = x^2$ at $x = 0$ for $x = 0.1$.

Answer: $L(x) = 0$. $f(0) = 0$, $f'(0) = 0$, so linearization gives zero.

Flashcard 3: What is $L(x)$ in the linear approximation formula?

Answer: The linear approximation of $f(x)$ near $x = a$. The linearization function that approximates $f(x)$.

Flashcard 4: Evaluate $L(x)$ for $f(x) = \text{sqrt}(x)$ at $x = 4$ using $x = 4.1$.

Answer: $L(x) = 2 + \frac{0.1}{4} = 2.025$. $f(4) = 2$, $f'(4) = \frac{1}{4}$, so $L(4.1) = 2 + \frac{1}{4}(0.1)$.

Flashcard 5: Find the linear approximation of $f(x) = \frac{1}{x}$ at $x = 1$ for $x = 0.9$.

Answer: $L(x) = 1 + 0.1 = 1.1$. $f(1) = 1$, $f'(1) = -1$, so $L(0.9) = 1 + (-1)(-0.1)$.

Flashcard 6: What is the linear approximation of $f(x) = e^x$ at $x = 1$?

Answer: $L(x) = e + e(x-1)$. $f(1) = e$, $f'(1) = e$, standard exponential linearization.

Flashcard 7: Calculate the linear approximation of $f(x) = x^2$ at $x=1$ for $x=1.1$.

Answer: $1 + 2(0.1) = 1.2$. $f(1) = 1$, $f'(1) = 2$, so $L(1.1) = 1 + 2(0.1)$.

Flashcard 8: Identify the derivative in the linearization formula $L(x) = f(a) + f'(a)(x-a)$.

Answer: $f'(a)$. This term represents the slope of the tangent line.

Flashcard 9: Find the linear approximation of $f(x) = \text{tan}(x)$ at $x = 0$ for $x = 0.05$.

Answer: $L(x) = 0.05$. $f(0) = 0$, $f'(0) = 1$, so $L(0.05) = 0 + 1(0.05)$.

Flashcard 10: What is the linear approximation of $f(x) = \text{exp}(x)$ at $x = 1$?

Answer: $L(x) = e + e(x-1)$. $f(1) = e$, $f'(1) = e$, so linearization has slope $e$.

Flashcard 11: Evaluate $L(x)$ for $f(x) = \frac{1}{x}$ at $x = 2$ using $x = 2.1$.

Answer: $L(x) = \frac{1}{2} - \frac{0.1}{4} = 0.475$. $f(2) = 0.5$, $f'(2) = -0.25$, so $L(2.1) = 0.5 - 0.25(0.1)$.

Flashcard 12: Approximate $\text{ln}(1.1)$ using linearization at $x = 1$ for $f(x) = \text{ln}(x)$.

Answer: $L(x) = 0 + 0.1 = 0.1$. $f(1) = 0$, $f'(1) = 1$, so $L(1.1) = 0 + 1(0.1)$.

Flashcard 13: Calculate the linear approximation of $f(x) = x^3$ at $x = 0$ for $x = 0.1$.

Answer: $L(x) = 0$. $f(0) = 0$, $f'(0) = 0$, so linearization gives zero.

Flashcard 14: Determine the linear approximation of $f(x) = x^4$ at $x = 1$ for $x = 1.1$.

Answer: $L(x) = 1 + 0.4 = 1.4$. $f(1) = 1$, $f'(1) = 4$, so $L(1.1) = 1 + 4(0.1)$.

Flashcard 15: Approximate $e^{0.2}$ using linearization at $x = 0$ for $f(x) = e^x$.

Answer: $L(x) = 1 + 0.2 = 1.2$. $f(0) = 1$, $f'(0) = 1$, so $L(0.2) = 1 + 1(0.2)$.

Flashcard 16: What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = 1$?

Answer: $L(x) = x - 1$. $f(1) = 0$, $f'(1) = 1$, giving the simple linear form.

Flashcard 17: Approximate $\text{sin}(0.1)$ using linearization at $x = 0$ for $f(x) = \text{sin}(x)$.

Answer: $L(x) = 0.1$. $f(0) = 0$, $f'(0) = 1$, so $L(0.1) = 0 + 1(0.1)$.

Flashcard 18: Determine the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$ for $x = e+0.1$.

Answer: $L(x) = 1 + 0.1/e$. $f(e) = 1$, $f'(e) = \frac{1}{e}$, so slope is $\frac{1}{e}$.

Flashcard 19: What is the linear approximation of $f(x) = \text{tan}(x)$ at $x = 0$?

Answer: For $x$ near $0$, $L(x) = x$. $f(0) = 0$, $f'(0) = \sec^2(0) = 1$, so linearization is $x$.

Flashcard 20: What is the role of the derivative in linear approximation?

Answer: It gives the slope of the tangent line. The derivative provides the rate of change for the approximation.

Flashcard 21: Determine the linear approximation of $f(x) = x^3$ at $x = 1$ for $x = 1.05$.

Answer: $L(x) = 1 + 3(0.05) = 1.15$. $f(1) = 1$, $f'(1) = 3$, so $L(1.05) = 1 + 3(0.05)$.

Flashcard 22: Find $L(x)$ for $f(x) = \text{ln}(x)$ at $x = 1$ when $x = 1.1$.

Answer: $L(x) = 0 + \frac{1}{1}(0.1) = 0.1$. $f(1) = 0$, $f'(1) = 1$, so $L(1.1) = 0 + 1(0.1)$.

Flashcard 23: Approximate $e^{0.1}$ using linearization at $x = 0$ for $f(x) = e^x$.

Answer: $L(x) = 1 + 0.1 = 1.1$. $f(0) = 1$, $f'(0) = 1$, so $L(0.1) = 1 + 1(0.1)$.

Flashcard 24: What does $f'(a)(x-a)$ represent in the linearization formula?

Answer: The change in the linear approximation. This represents how much the function changes linearly.

Flashcard 25: Which function value does $f(a)$ represent in the linear approximation formula?

Answer: The value of the function at $x = a$. This is the y-coordinate where the tangent line touches the curve.

Flashcard 26: What is the geometric interpretation of a linearization at $x = a$?

Answer: It's the tangent line to the function at $x = a$. The linearization creates the tangent line at that specific point.

Flashcard 27: Define local linearity in the context of a differentiable function.

Answer: Local linearity means the function appears linear near a point. The function's graph looks like a straight line when zoomed in close.

Flashcard 28: What is the approximated change in $f(x)$ for $f(x) = x^3$ at $x = 1$?

Answer: $3(x-1)$. For $f(x) = x^3$ at $x=1$, $f'(1) = 3$ gives the linear change.

Flashcard 29: What is the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$?

Answer: For $x$ near $0$, $L(x) = x$. $f(0) = 0$, $f'(0) = 1$, so linearization is just $x$.

Flashcard 30: For which type of function can local linearity be used to approximate values?

Answer: Differentiable functions. Must have a derivative to calculate the tangent line slope.