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AP Calculus BC Practice Quiz: Approximating Values of a Function Using Local Linearity and Linearization

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

Let f be a differentiable function such that f(2) = 5 and f'(2) = -3. What is the approximation for f(2.1) found by using the line tangent to the graph of f at x = 2?

All Questions (9)

Let f be a differentiable function such that f(2) = 5 and f'(2) = -3. What is the approximation for f(2.1) found by using the line tangent to the graph of f at x = 2?

A) 4.7

B) 4.9

C) 5.0

D) 5.3

Correct Answer: A

The equation of the tangent line (local linearization) is given by L(x) = f(a) + f'(a)(x-a). Here, a=2, f(2)=5, and f'(2)=-3. So, L(x) = 5 + (-3)(x-2). To approximate f(2.1), we calculate L(2.1) = 5 - 3(2.1 - 2) = 5 - 3(0.1) = 5 - 0.3 = 4.7.

The local linear approximation of a function f at x=c is the equation of the tangent line at that point. This concept is based on the idea that:

A) all functions are secretly lines if you look closely enough.

B) the tangent line's accuracy as an approximation improves as one moves further from the point of tangency.

C) near the point of tangency, a differentiable function's graph closely resembles its tangent line.

D) a tangent line approximation is only useful for functions that are concave up.

Correct Answer: C

This statement correctly describes the principle of local linearity. The tangent line is the graph of the locally linear approximation because, for a small interval around the point of tangency, the curve of the function is well-approximated by the straight line of the tangent.

Let f be a twice-differentiable function. The value of f(4.1) is approximated using the tangent line to the graph of f at x = 4. If f(4) = 2, f'(4) = 5, and f''(4) = -3, which of the following statements is true?

A) The approximation is 2.5, which is an overestimate.

B) The approximation is 2.5, which is an underestimate.

C) The approximation is 1.5, which is an overestimate.

D) The approximation is 1.5, which is an underestimate.

Correct Answer: A

First, find the approximation: L(4.1) = f(4) + f'(4)(4.1 - 4) = 2 + 5(0.1) = 2.5. Next, determine if it's an overestimate or underestimate using the second derivative. Since f''(4) = -3 < 0, the graph of f is concave down at x=4. For a concave down function, the tangent line lies above the curve, making the approximation an overestimate.

Use the line tangent to the graph of f(x) = √x at x = 9 to approximate the value of f(9.2).

A) 3.017

B) 3.033

C) 3.067

D) 3.100

Correct Answer: B

First, find the point and the slope. The point is (9, f(9)) = (9, 3). The derivative is f'(x) = 1/(2√x). The slope at x=9 is f'(9) = 1/(2√9) = 1/6. The tangent line is L(x) = 3 + (1/6)(x-9). The approximation is L(9.2) = 3 + (1/6)(9.2 - 9) = 3 + (1/6)(0.2) = 3 + 0.2/6 = 3 + 2/60 = 3 + 1/30 ≈ 3.033.

The graph of a twice-differentiable function g is shown. A tangent line is drawn at x = 2. Based on the graph, the tangent line approximation for g(1.9) would be:

A) an overestimate, because the graph of g is increasing at x=2.

B) an underestimate, because the graph of g is increasing at x=2.

C) an overestimate, because the graph of g is concave down at x=2.

D) an underestimate, because the graph of g is concave up at x=2.

Correct Answer: D

The graph at x=2 is bending upwards, which indicates it is concave up. For a function that is concave up, the tangent line lies below the curve. Therefore, any approximation using the tangent line will be an underestimate of the actual function value. Whether the function is increasing or decreasing does not determine if the approximation is an over- or underestimate.

A function f is differentiable and has values given in the table. What is the estimate for f(2.8) using the local linear approximation of f at x = 3? | x | f(x) | f'(x) | |---|---|---| | 2 | 8 | 4 | | 3 | 6 | -3 |

A) 5.4

B) 5.7

C) 6.3

D) 6.6

Correct Answer: D

We use the tangent line at x=3. From the table, f(3) = 6 and f'(3) = -3. The local linear approximation is L(x) = f(3) + f'(3)(x-3) = 6 - 3(x-3). We want to estimate f(2.8), so we calculate L(2.8) = 6 - 3(2.8 - 3) = 6 - 3(-0.2) = 6 + 0.6 = 6.6.

Consider the curve defined by the equation x³ + y³ = 9. Use the tangent line to the curve at the point (1, 2) to approximate the y-coordinate of the point on the curve where x = 0.9.

A) 1.925

B) 2.025

C) 2.075

D) 2.125

Correct Answer: B

First, find dy/dx using implicit differentiation: 3x² + 3y²(dy/dx) = 0, so dy/dx = -x²/y². At the point (1, 2), the slope is dy/dx = -(1)²/(2)² = -1/4. The tangent line equation is y - 2 = (-1/4)(x - 1). The approximation is L(x) = 2 - (1/4)(x-1). For x=0.9, L(0.9) = 2 - (1/4)(0.9 - 1) = 2 - (1/4)(-0.1) = 2 + 0.1/4 = 2 + 0.025 = 2.025.

Let g be the function given by g(x) = ∫[2, x] ln(t² + 1) dt. Use the local linear approximation of g at x = 2 to estimate g(2.1).

A) 0

B) ln(5)

C) 0.1 * ln(5)

D) 0.1 + ln(5)

Correct Answer: C

We need g(2) and g'(2). First, g(2) = ∫[2, 2] ln(t² + 1) dt = 0. By the Fundamental Theorem of Calculus, Part 1, g'(x) = ln(x² + 1). So, g'(2) = ln(2² + 1) = ln(5). The linear approximation is L(x) = g(2) + g'(2)(x-2) = 0 + ln(5)(x-2). The estimate for g(2.1) is L(2.1) = ln(5)(2.1 - 2) = 0.1 * ln(5).

Let f be a function with f'(x) = e^(x²) and f(0) = 5. The tangent line to the graph of f at x = 0 is used to approximate f(0.1). Which of the following statements is true?

A) The approximation is 5.1, and it is an overestimate.

B) The approximation is 5.1, and it is an underestimate.

C) The approximation is 6.0, and it is an overestimate.

D) The approximation is 6.0, and it is an underestimate.

Correct Answer: B

The point of tangency is (0, 5). The slope is f'(0) = e^(0²) = 1. The approximation is L(0.1) = f(0) + f'(0)(0.1 - 0) = 5 + 1(0.1) = 5.1. To determine if it is an over- or underestimate, we find the second derivative: f''(x) = d/dx(e^(x²)) = e^(x²) * 2x. At x=0, f''(0) = 0, which is inconclusive. However, for x in the interval (0, 0.1), f''(x) = 2x * e^(x²) is positive. Since f''(x) > 0 on the interval, the function is concave up. Therefore, the tangent line lies below the curve, and the approximation is an underestimate.