AP Calculus AB 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 AB.

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 approximation $L(x)$ when $f(x) = \text{tan}(x)$ at $x = 0$?

Answer: $L(x) = x$. Since $\tan(0) = 0$ and $\tan'(0) = \sec^2(0) = 1$.

Flashcard 2: What is the approximation $L(x)$ when $f(x) = \text{tan}(x)$ at $x = 0$?

Answer: $L(x) = x$. Since $\tan(0) = 0$ and $\tan'(0) = \sec^2(0) = 1$.

Flashcard 3: Identify the derivative required for linearization of $f(x)$ at $x = a$.

Answer: $f'(a)$. The slope of the tangent line at the linearization point.

Flashcard 4: Which function value is used in the linear approximation $L(x) = f(a) + f'(a)(x - a)$?

Answer: $f(a)$. The y-intercept when the tangent line passes through $(a, f(a))$.

Flashcard 5: Determine the linearization of $f(x) = x^2$ at $x = 2$.

Answer: $L(x) = 4 + 4(x - 2)$. Since $f(2) = 4$ and $f'(x) = 2x$, so $f'(2) = 4$.

Flashcard 6: Identify the linear approximation for $f(x) = \text{exp}(x)$ at $x = 1$.

Answer: $L(x) = e + e(x - 1)$. Since $e^1 = e$ and $(e^x)' = e^x$, so $f'(1) = e$.

Flashcard 7: What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = 1$?

Answer: $L(x) = x - 1$. Since $\ln(1) = 0$ and $(\ln x)' = \frac{1}{x}$, so $f'(1) = 1$.

Flashcard 8: Which point is used as the center for a linear approximation?

Answer: The point $a$ where the approximation is centered. The base point where the tangent line touches the curve.

Flashcard 9: Identify the linear approximation of $f(x) = \text{cos}(x)$ at $x = 0$.

Answer: $L(x) = 1$. Since $\cos(0) = 1$ and $\cos'(0) = -\sin(0) = 0$.

Flashcard 10: Which function approximates $f(x)$ near $x = a$ using local linearity?

Answer: The linearization $L(x)$ of $f(x)$. The best linear approximation near the given point.

Flashcard 11: Find the linear approximation of $f(x) = \text{arcsin}(x)$ at $x = 0$.

Answer: $L(x) = x$. Since $\arcsin(0) = 0$ and $(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}$.

Flashcard 12: Calculate the derivative needed for linearization of $f(x) = \frac{1}{x}$ at $x = 1$.

Answer: $f'(x) = -\frac{1}{x^2}$, so $f'(1) = -1$. Using the power rule and evaluating at $x = 1$.

Flashcard 13: What is the linear approximation of $f(x) = \text{arctan}(x)$ at $x = 0$?

Answer: $L(x) = x$. Since $\arctan(0) = 0$ and $(\arctan x)' = \frac{1}{1+x^2}$.

Flashcard 14: What is the linear approximation of $f(x) = e^x$ at $x = 0$?

Answer: $L(x) = 1 + x$. Since $e^0 = 1$ and $(e^x)' = e^x$, so $f'(0) = 1$.

Flashcard 15: State the condition under which a function is locally linear at a point.

Answer: The derivative $f'(x)$ exists and is continuous near the point. Ensures the function behaves like a line near that point.

Flashcard 16: Find the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$.

Answer: $L(x) = x$. Since $\sin(0) = 0$ and $\sin'(0) = \cos(0) = 1$.

Flashcard 17: Calculate the derivative needed for linearization of $f(x) = \frac{1}{x}$ at $x = 1$.

Answer: $f'(x) = -\frac{1}{x^2}$, so $f'(1) = -1$. Using the power rule and evaluating at $x = 1$.

Flashcard 18: What is the linear approximation of $f(x) = x^4$ at $x = 1$?

Answer: $L(x) = 1 + 4(x - 1)$. Since $f(1) = 1$ and $f'(x) = 4x^3$, so $f'(1) = 4$.

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

Answer: $L(x) = x$. Since $\arctan(0) = 0$ and $(\arctan x)' = \frac{1}{1+x^2}$.

Flashcard 20: State the condition under which a function is locally linear at a point.

Answer: The derivative $f'(x)$ exists and is continuous near the point. Ensures the function behaves like a line near that point.

Flashcard 21: Find the linear approximation of $f(x) = \text{sin}(x)$ at $x = 0$.

Answer: $L(x) = x$. Since $\sin(0) = 0$ and $\sin'(0) = \cos(0) = 1$.

Flashcard 22: Identify the derivative required for linearization of $f(x)$ at $x = a$.

Answer: $f'(a)$. The slope of the tangent line at the linearization point.

Flashcard 23: What is the formula for the linear approximation of a function at point $a$?

Answer: $L(x) = f(a) + f'(a)(x - a)$. This is the tangent line equation at point $a$.

Flashcard 24: What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = 1$?

Answer: $L(x) = x - 1$. Since $\ln(1) = 0$ and $(\ln x)' = \frac{1}{x}$, so $f'(1) = 1$.

Flashcard 25: Which point is used as the center for a linear approximation?

Answer: The point $a$ where the approximation is centered. The base point where the tangent line touches the curve.

Flashcard 26: Find the linear approximation of $f(x) = \text{arcsin}(x)$ at $x = 0$.

Answer: $L(x) = x$. Since $\arcsin(0) = 0$ and $(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}$.

Flashcard 27: What is the linear approximation for $f(x) = \text{sqrt}(x)$ at $x = 4$?

Answer: $L(x) = 2 + \frac{1}{4}(x - 4)$. Since $\sqrt{4} = 2$ and $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$.

Flashcard 28: State the linear approximation for $f(x) = \frac{1}{x}$ at $x = 2$.

Answer: $L(x) = \frac{1}{2} - \frac{1}{4}(x - 2)$. Since $f(2) = \frac{1}{2}$ and $f'(2) = -\frac{1}{4}$.

Flashcard 29: Identify the linear approximation for $f(x) = \text{exp}(x)$ at $x = 1$.

Answer: $L(x) = e + e(x - 1)$. Since $e^1 = e$ and $(e^x)' = e^x$, so $f'(1) = e$.

Flashcard 30: What is the linear approximation of $f(x) = x^5$ at $x = 1$?

Answer: $L(x) = 1 + 5(x - 1)$. Since $f(1) = 1$ and $f'(x) = 5x^4$, so $f'(1) = 5$.