Home / AP Calculus AB : 2.2 Defining the Derivative of a Function and Using Derivative Notation – Exam Style questions with Answer- MCQ

AP Calculus AB : 2.2 Defining the Derivative of a Function and Using Derivative Notation – Exam Style questions with Answer- MCQ

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Question

Graph of f'(x)

The graph of the even function \( y = f'(x) \) consists of 4 line segments, as shown above. Which of the following statements about \( f \) is false?

A) \( \lim_{x\rightarrow 0}(f(x)-f(0))=0 \)
B) \( \lim_{x\rightarrow 0}\frac{f(x)-f(-x)}{x}=0 \)
C) \( \lim_{x\rightarrow 0} \frac{f(x)-f(-x)}{2x}=0 \)
D) \( \lim_{x\rightarrow 2}\frac{f(x)-f(2)}{x-2}=1 \)
E) \( \lim_{x\rightarrow 3}\frac{f(x)-f(3)}{x-3} \) does not exist

▶️ Answer/Explanation

Solution

Correct Answer: B

Since \( f’ \) is even, \( f \) is not necessarily even. Statement B resembles the derivative of an odd function, and is not guaranteed to be true. Therefore, statement B is false.

Question

Consider the piecewise function defined by: \( f(x) = \begin{cases} 2x – 2 & \text{for } x < 3 \\ 2x – 4 & \text{for } x \geq 3 \end{cases} \)
Which of the following statements about this function is true?

(A) The function is continuous at x = 3
(B) The limit as x approaches 3 exists
(C) The function has a removable discontinuity at x = 3
(D) The function has a jump discontinuity at x = 3
(E) The function is differentiable at x = 3

▶️ Answer/Explanation

Solution

Correct Answer: D

1. Evaluate the function at x = 3:
– Left approach (x→3⁻): \( f(3^-) = 2(3) – 2 = 4 \)
– Right approach (x→3⁺): \( f(3^+) = 2(3) – 4 = 2 \)
– Actual value: \( f(3) = 2(3) – 4 = 2 \)

2. Analyze continuity and differentiability:
– The left and right limits are different (4 ≠ 2)
– The function value equals the right limit (f(3) = 2)
– This creates a jump discontinuity at x = 3

Key observations:
• (A) False – Not continuous (left ≠ right limit)
• (B) False – Limit doesn’t exist (left ≠ right)
• (C) False – Not removable (finite jump)
• (D) True – Jump discontinuity present
• (E) False – Not differentiable (not continuous)

Question

The derivative of a function f is given by \( f'(x) = 0.2x + e^{0.15x} \). Which of the following procedures can be used to determine the value of x at which the line tangent to the graph of f has slope 2?

A) Evaluate \( 0.2x + e^{0.15x} \) at x=2
B) Evaluate \( \frac{d}{dx}(0.2x + e^{0.15x}) \) at x=2
C) Solve \( 0.2x + e^{0.15x} = 2 \) for x
D) Solve \( \frac{d}{dx}(0.2x + e^{0.15x}) = 2 \) for x

▶️ Answer/Explanation

Solution

Correct Answer: C

Key Concept:
The derivative \( f'(x) \) represents the slope of the tangent line to the graph of \( f \) at any point \( x \).

Solution Steps:
1. We need to find where the tangent slope equals 2
2. Set the derivative equal to 2: \( 0.2x + e^{0.15x} = 2 \)
3. Solve this equation for \( x \) (option C)

Note: The equation \( 0.2x + e^{0.15x} = 2 \) would typically be solved numerically or graphically, as it cannot be solved algebraically.

Question

The derivative of the function f is given by \( f'(x) = -2x + 4 \) for all x, and \( f(-1) = 5 \). Which of the following is an equation for the line tangent to the graph of f at \( x = -1 \)?

A) \( y = -2x + 3 \)
B) \( y = -2x + 4 \)
C) \( y = 6x + 5 \)
D) \( y = 6x + 11 \)

▶️ Answer/Explanation

Solution

Correct Answer: D

Step 1: Find the slope at \( x = -1 \)
\( f'(-1) = -2(-1) + 4 = 2 + 4 = 6 \)

Step 2: Use the given point \( (-1, 5) \)
We know the tangent line passes through this point with slope 6.

Step 3: Write the equation in point-slope form
\( y – 5 = 6(x – (-1)) \)
\( y = 6(x + 1) + 5 \)

Step 4: Simplify to slope-intercept form
\( y = 6x + 6 + 5 \)
\( y = 6x + 11 \)

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