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: Evaluate $L(x)$ for $f(x) = \text{cos}(x)$ at $x = \frac{\text{π}}{2}$ using $x = \frac{\text{π}}{2} + 0.1$.

Answer: $L(x) = 0 - 0.1 = -0.1$. $f(\frac{\pi}{2}) = 0$, $f'(\frac{\pi}{2}) = -1$, so slope is negative.

Flashcard 2: 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 3: 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 4: 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 5: What is the formula for the linear approximation of a function $f$ at $x=a$?

Answer: $L(x) = f(a) + f'(a)(x-a)$. Standard formula where $f(a)$ is the point and $f'(a)$ is the slope.

Flashcard 6: 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 7: 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 8: 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 9: 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 10: 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 11: What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$?

Answer: $L(x) = 1 + \frac{x-e}{e}$. $f(e) = 1$, $f'(e) = \frac{1}{e}$, standard logarithm linearization.

Flashcard 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: Determine $L(x)$ for $f(x) = e^x$ at $x = 0$ with $x = 0.05$.

Answer: $L(x) = 1 + 0.05 = 1.05$. $f(0) = 1$, $f'(0) = 1$, so $L(0.05) = 1 + 1(0.05)$.

Flashcard 27: 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 28: What is the linear approximation of $f(x) = \text{ln}(x)$ at $x = e$?

Answer: $L(x) = 1 + \frac{x-e}{e}$. $f(e) = 1$, $f'(e) = \frac{1}{e}$, standard logarithm linearization.

Flashcard 29: 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 30: 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 31: 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 32: 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 33: 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 34: 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 35: 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 36: 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 37: 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.

Flashcard 38: 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 39: 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 40: 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 41: 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 42: 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 43: 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.

Flashcard 44: 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 45: 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 46: Determine $L(x)$ for $f(x) = e^x$ at $x = 0$ with $x = 0.05$.

Answer: $L(x) = 1 + 0.05 = 1.05$. $f(0) = 1$, $f'(0) = 1$, so $L(0.05) = 1 + 1(0.05)$.

Flashcard 47: 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 48: 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 49: 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 50: 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 51: 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 52: 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 53: 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 54: 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 55: 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 56: 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 57: 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 58: 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 59: Evaluate $L(x)$ for $f(x) = \text{cos}(x)$ at $x = \frac{\text{π}}{2}$ using $x = \frac{\text{π}}{2} + 0.1$.

Answer: $L(x) = 0 - 0.1 = -0.1$. $f(\frac{\pi}{2}) = 0$, $f'(\frac{\pi}{2}) = -1$, so slope is negative.

Flashcard 60: 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 61: 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 62: 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 63: 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 64: 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 65: What is the formula for the linear approximation of a function $f$ at $x=a$?

Answer: $L(x) = f(a) + f'(a)(x-a)$. Standard formula where $f(a)$ is the point and $f'(a)$ is the slope.

Flashcard 66: 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 67: 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 68: 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 $L(0.05) = 0 + 1(0.05)$.