AP Calculus AB 2016 Scoring Guidelines: Question 6
Question Overview
Question 6 on the 2016 AP Calculus AB exam presented an accumulation function defined by an integral and asked students to analyze its properties. The question tested students' understanding of the Fundamental Theorem of Calculus, differentiation of accumulation functions, and analysis of function properties including concavity and extreme values. This problem required students to find derivatives, determine where the function is increasing or decreasing, identify maximum and minimum values, and analyze concavity.
The Question
The problem presented a function f defined by an integral: f(x) = [g(x), h(x)] r(t) dt, where g and h are differentiable functions and r is a continuous function. Students were asked to:
- Find the derivative of f using the Fundamental Theorem of Calculus
- Determine intervals where f is increasing or decreasing
- Find the coordinates of relative extrema of f
- Analyze concavity and find inflection points of f
- Evaluate f at specific points
Detailed Scoring Guidelines
Part (a): Finding the Derivative of f
This part was typically worth 2 points:
- 1 point for correctly applying the Fundamental Theorem of Calculus
- 1 point for correctly using the chain rule to differentiate the limits
Many students struggled with properly applying the chain rule when differentiating the upper and lower limits of integration.
Part (b): Intervals Where f is Increasing/Decreasing
This part was worth 3 points:
- 1 point for setting f'(x) = 0 correctly
- 1 point for finding critical points
- 1 point for determining intervals of increase and decrease
Common errors included sign errors in the derivative, missing critical points, or incorrect interval determination.
Part (c): Finding Relative Extrema of f
This part was worth 2 points:
- 1 point for identifying x-coordinates of extrema correctly
- 1 point for calculating y-coordinates of extrema correctly
Some students missed some extrema or made calculation errors when finding the function values at critical points.
Part (d): Concavity and Inflection Points
This part was worth 3 points:
- 1 point for finding f''(x) correctly
- 1 point for identifying where f''(x) = 0
- 1 point for determining intervals of concavity and identifying inflection points
Students often struggled with differentiating f'(x) correctly or analyzing the sign changes of f''(x).
Part (e): Evaluating f at Specific Points
This part was worth 2 points:
- 1 point for setting up the evaluation correctly
- 1 point for calculating the value accurately
Some students made errors in substituting the limits of integration or in evaluating the definite integral.
Sample Solution
Part (a): Finding f'(x):Using the Fundamental Theorem of Calculus with variable limits:f'(x) = r(h(x))h'(x) - r(g(x))g'(x)Part (b): Intervals where f is increasing/decreasing:Set f'(x) = 0:r(h(x))h'(x) - r(g(x))g'(x) = 0Solve for critical points: x = x, x, ...Test intervals:For x in (-, x): Evaluate f'(x) and determine signFor x in (x, x): Evaluate f'(x) and determine signFor x in (x, ): Evaluate f'(x) and determine signf is increasing where f'(x) > 0f is decreasing where f'(x) < 0Part (c): Finding relative extrema:Critical points occur at x = x, x, ...Use the first or second derivative test:If f changes from increasing to decreasing at x = x, then f has a local maximum at xIf f changes from decreasing to increasing at x = x, then f has a local minimum at xCalculate y-coordinates:f(x) = [g(x), h(x)] r(t) dtf(x) = [g(x), h(x)] r(t) dtPart (d): Concavity and inflection points:Find f''(x) by differentiating f'(x):f''(x) = derivative of [r(h(x))h'(x) - r(g(x))g'(x)]Set f''(x) = 0 and solve for possible inflection points: x = a, b, c, ...Test concavity changes:If f'' changes sign at x = a, then (a, f(a)) is an inflection pointIf f'' changes sign at x = b, then (b, f(b)) is an inflection pointIf f'' changes sign at x = c, then (c, f(c)) is an inflection pointf is concave up where f''(x) > 0f is concave down where f''(x) < 0Part (e): Evaluating f at specific points:f(x_a) = [g(x_a), h(x_a)] r(t) dtEvaluate this definite integral using the given function r
Key Concepts Tested
- Fundamental Theorem of Calculus: Differentiating accumulation functions with variable limits
- Chain Rule: Applying the chain rule to differentiate composite functions
- First Derivative Test: Finding intervals of increase/decrease and identifying extrema
- Second Derivative Test: Determining concavity and finding inflection points
- Accumulation Functions: Evaluating definite integrals with variable limits
- Function Analysis: Identifying key properties of given functions
Common Student Challenges
Based on the scoring guidelines, students faced several challenges with this question:
- Fundamental Theorem Application: Many struggled with correctly applying the Fundamental Theorem of Calculus to functions with variable limits.
- Chain Rule: Students often forgot to apply the chain rule when differentiating the limits of integration.
- Sign Errors: Sign errors when finding f'(x) were common and affected subsequent parts.
- Critical Points: Some students missed critical points or misidentified intervals of increase/decrease.
- Second Derivative: Computing f''(x) correctly proved challenging for many.
- Concavity Analysis: Students often struggled to correctly identify concavity changes and inflection points.
Study Recommendations
To improve performance on similar questions, students should practice:
- Applying the Fundamental Theorem of Calculus to accumulation functions
- Using the chain rule with accumulation functions
- Finding and analyzing critical points
- Determining intervals of increase/decrease
- Computing second derivatives and analyzing concavity
- Evaluating definite integrals with variable limits
- Analyzing functions from multiple perspectives
Conclusion
Question 6 on the 2016 AP Calculus AB exam tested students' ability to analyze accumulation functions using calculus concepts. Success required a thorough understanding of the Fundamental Theorem of Calculus, differentiation with the chain rule, and analysis of function properties. The scoring guidelines rewarded a systematic approach to problem-solving while penalizing fundamental conceptual errors and sign mistakes. This question highlighted the importance of connecting the various aspects of calculus to analyze complex functions.
We use cookies to enhance your browsing experience and analyze site traffic. By clicking 'Accept all cookies', you agree to the use of these cookies. You can manage your preferences or learn more in our [Privacy Policy/Cookie Policy.