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: 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: 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 9: 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 10: Determine the linear approximation of $f(x) = x^3$ at $x = 1$.

Answer: $L(x) = 1 + 3(x - 1)$. Since $f(1) = 1$ and $f'(x) = 3x^2$, so $f'(1) = 3$.

Flashcard 11: 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 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: Determine the linear approximation of $f(x) = x^3$ at $x = 1$.

Answer: $L(x) = 1 + 3(x - 1)$. Since $f(1) = 1$ and $f'(x) = 3x^2$, so $f'(1) = 3$.

Flashcard 14: 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 15: 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 16: 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 17: 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 18: 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 19: 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 20: What role does $f'(a)$ play in linearization?

Answer: It is the slope of the tangent line at $x = a$. It determines the steepness of the linear approximation.

Flashcard 21: 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 22: 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 23: 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 24: 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 $f'(1) = 5$.

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: 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 27: 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 28: 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 29: 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 30: What role does $f'(a)$ play in linearization?

Answer: It is the slope of the tangent line at $x = a$. It determines the steepness of the linear approximation.

Flashcard 31: 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 32: 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 33: 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 34: 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 35: 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 36: 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 37: 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 38: 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 39: What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$?

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

Flashcard 40: What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$?

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

Flashcard 41: 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 $f'(1) = 5$.

Flashcard 42: 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 43: 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 44: 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 45: Identify the linear approximation of $f(x) = \text{cos}(x)$ at $x = 0$.

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