AP Calculus AB 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
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Question 1 of 9
All Questions (9)
A) 4.8
B) 4.9
C) 5.0
D) 5.2
Correct Answer: A
The equation of the tangent line (local linear approximation) at x=a is L(x) = f(a) + f'(a)(x-a). Here, a=3, f(3)=5, and f'(3)=-2. So, L(x) = 5 - 2(x-3). To approximate f(3.1), we calculate L(3.1) = 5 - 2(3.1 - 3) = 5 - 2(0.1) = 5 - 0.2 = 4.8.
A) The approximation for f(5.1) is an underestimate.
B) The approximation for f(5.1) is an overestimate.
C) The approximation for f(5.1) is exact.
D) It cannot be determined if the approximation is an overestimate or an underestimate.
Correct Answer: B
The condition f''(x) < 0 means that the graph of f is concave down. When a function is concave down, its tangent line lies above the curve. Therefore, any approximation using the tangent line will be greater than the actual function value, resulting in an overestimate.
A) The Mean Value Theorem
B) The Intermediate Value Theorem
C) The local linear approximation of the function
D) The global behavior of the function
Correct Answer: C
The tangent line provides a good approximation for a function's value near the point of tangency because the function is 'locally linear'. This means that if you zoom in sufficiently on the point, the curve and its tangent line become nearly indistinguishable. The tangent line is the graph of this local linear approximation.
A) 2.9667
B) 2.9600
C) 3.0333
D) 3.0400
Correct Answer: A
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 equation is y - 3 = (1/6)(x - 9). The approximation is L(8.8) = 3 + (1/6)(8.8 - 9) = 3 + (1/6)(-0.2) = 3 - 0.2/6 = 3 - 1/30 ≈ 3 - 0.0333 = 2.9667.
A) g(2) = 11 and g'(2) = -4
B) g(2) = 3 and g'(2) = -4
C) g(2) = -4 and g'(2) = 11
D) g(2) = -4 and g'(2) = 3
Correct Answer: B
The formula for local linear approximation is L(x) = g(a) + g'(a)(x-a). For a=2, this is L(x) = g(2) + g'(2)(x-2). Expanding this gives L(x) = g'(2)x + (g(2) - 2g'(2)). We are given L(x) = -4x + 11. By comparing the coefficients of x, we see that g'(2) = -4. By comparing the constant terms, we have g(2) - 2g'(2) = 11. Substituting g'(2) = -4 gives g(2) - 2(-4) = 11, which simplifies to g(2) + 8 = 11, so g(2) = 3.
A) The approximation is -0.9, and it is an underestimate.
B) The approximation is -0.9, and it is an overestimate.
C) The approximation is -1.1, and it is an underestimate.
D) The approximation is -1.1, and it is an overestimate.
Correct Answer: A
First, find the approximation. f(1) = 1^3 - 2(1) = -1. f'(x) = 3x^2 - 2, so f'(1) = 3(1)^2 - 2 = 1. The tangent line is y - (-1) = 1(x - 1), or y = x - 2. The approximation for f(1.1) is L(1.1) = 1.1 - 2 = -0.9. Next, determine if it's an over/underestimate by checking concavity. f''(x) = 6x. At x=1, f''(1) = 6 > 0, so the function is concave up. A tangent to a concave up function lies below the curve, making the approximation an underestimate.
A) 3.7
B) 3.9
C) 4.1
D) 4.3
Correct Answer: A
The equation of the tangent line is given, which is the local linear approximation. To estimate f(1.9), simply substitute x = 1.9 into the equation: y = 3(1.9 - 2) + 4 = 3(-0.1) + 4 = -0.3 + 4 = 3.7.
A) f(2.1) < L(2.1) < f(2)
B) L(2.1) < f(2.1) < f(2)
C) f(2) < f(2.1) < L(2.1)
D) f(2) < L(2.1) < f(2.1)
Correct Answer: B
Since f is concave up, the tangent line lies below the curve, which means the approximation L(2.1) is an underestimate of the actual value f(2.1). Thus, L(2.1) < f(2.1). Since f is decreasing on the interval, its values decrease as x increases. Therefore, f(2.1) < f(2). Combining these two inequalities gives the relationship L(2.1) < f(2.1) < f(2).
A) Approximation is 4.4; it is an overestimate.
B) Approximation is 4.4; it is an underestimate.
C) Approximation is 5.6; it is an overestimate.
D) Approximation is 5.6; it is an underestimate.
Correct Answer: B
The tangent line approximation at x=1 is L(x) = f(1) + f'(1)(x-1). Using the table values, L(x) = 5 + (-3)(x-1). The approximation for f(1.2) is L(1.2) = 5 - 3(1.2 - 1) = 5 - 3(0.2) = 5 - 0.6 = 4.4. To determine if this is an over- or underestimate, we look at the sign of the second derivative. At x=1, f''(1) = 4, which is positive. A positive second derivative indicates the function is concave up. For a concave up function, the tangent line lies below the curve, so the approximation is an underestimate.