2017 AP Calculus AB Free Response #3
TLDRIn this video, Alan from Bothell STEM, a coach, guides viewers through AP Calculus 2017 free response question number three, focusing on the non-calculator portion. The video explains how to find the values of F(-6) and F(5) by using the fundamental theorem of calculus to compute the antiderivative of the given function's derivative. Alan also discusses how to determine the intervals on which the function is increasing and how to find the absolute minimum value of the function on the closed interval. Additionally, the video touches on the concept of the second derivative, explaining why it exists at certain points and does not at others. The video concludes with a review of the scoring guidelines for the question.
Takeaways
- π Alan is teaching AP Calculus and working on question number three from the 2017 free response section.
- βοΈ The question requires solving without a calculator, meaning all calculations must be done by hand.
- π The function f is differentiable on the closed interval from -6 to 5 and satisfies a given equation.
- π Alan uses the graph of the derivative of f (f') to find the values of f(-6) and f(5) by integrating the derivative.
- π’ The value of f(-6) is calculated to be 3 and f(5) is found to be 10 - 2Ο using the fundamental theorem of calculus.
- π The intervals on which the function f is increasing are identified as being between -6 and -2 and from 2 to 5.
- π To find the absolute minimum value of f, Alan evaluates the function at critical points and endpoints of the interval.
- π The absolute minimum value of f on the closed interval is determined to be at f(2), which is 7 - 2Ο.
- π€ For part D, Alan calculates the second derivative at x = -5, which is -1/2, and notes that it does not exist at x = 3 due to differing left and right limits.
- π Alan explains the scoring guidelines for the question, noting that precision in notation is not strictly required.
- π The video concludes with an invitation for viewers to comment, like, subscribe, and a promise to continue with the next question in a future video.
Q & A
What is the main topic of the video?
-The video is about solving AP Calculus 2017 free response question number three, which involves finding values of a function F on a closed interval and understanding its behavior without the use of a calculator.
What is the relationship between F and F' (the derivative of F)?
-The relationship between F and F' is that F is the antiderivative of the derivative F'. This is based on the Fundamental Theorem of Calculus, which states that the integral (antiderivative) of a derivative is the original function up to a constant.
How does Alan determine the value of F at negative 6 and 5?
-Alan determines the value of F at negative 6 and 5 by integrating the given derivative function from negative 2 to these points and then adjusting by the initial value of F at negative 2, which is 7.
What is the area under the curve from negative 2 to negative 6 in the context of the integral?
-The area under the curve from negative 2 to negative 6 is considered negative because the integration is from right to left. It consists of a triangle with a height of 2 and a base of 4, resulting in an area of 4, which when accounted for the direction of integration, becomes -4.
What is the value of F at negative 6?
-The value of F at negative 6 is 3, which is obtained by adding the initial value F(-2) which is 7, and then subtracting the negative area under the curve from -2 to -6, which is 4.
How does Alan find the intervals on which the function F is increasing?
-Alan finds the intervals where F is increasing by determining when the derivative F' is greater than zero, which corresponds to the graph of F' being above the x-axis.
What are the intervals on which the function F is increasing?
-The function F is increasing on the intervals (-6, -2) and (2, 5), not including the point -2 where the derivative is zero.
How does Alan determine the absolute minimum value of F on the closed interval?
-Alan determines the absolute minimum value of F by evaluating the function at critical points where the derivative is zero and at the endpoints of the interval, then comparing these values to find the smallest one.
What is the value of the second derivative of F at negative 5?
-The second derivative of F at negative 5 is -1/2, which is the slope of the tangent to the graph of the first derivative at that point.
Why doesn't the second derivative of F exist at 3?
-The second derivative of F does not exist at 3 because the left and right limits of the slope as x approaches 3 are not the same, indicating an abrupt change in the slope at that point.
What is the minimum value of F on the closed interval according to the video?
-The minimum value of F on the closed interval is at the point where F(2) equals 7 minus 2Ο, which is approximately less than 1, making it the smallest value among the evaluated points.
Outlines
π AP Calculus 2017 Question 3: Non-Calculator Portion
In this segment, Alan from Bothell Stem introduces the third free response question from the 2017 AP Calculus exam. The question involves a function f that is differentiable on the interval from -6 to 5 and satisfies a given equation. Alan explains that f is the antiderivative of its derivative, f'. He uses the fundamental theorem of calculus to find the values of f at -6 and 5 by integrating the derivative function from -2 to these points and adjusting for the initial condition f(-2) = 7. The area under the curve is calculated to determine f(-6) and f(5), with the latter involving a semicircle and a triangle. The explanation also covers how to determine intervals of increase for the function and concludes with the solution to part A of the question.
π Finding the Absolute Minimum Value of f
Alan continues with part B of the question, focusing on finding the absolute minimum value of the function f on the closed interval. He explains the process of identifying critical numbers where the derivative is zero or undefined. Since the derivative is defined everywhere, Alan looks for points where it equals zero, which are at x = -2, 2, and 5. He also considers the endpoints of the interval, -6 and 5. After evaluating f at these points, he determines that the smallest value occurs at x = 2, with f(2) = 7 - 2Ο. Alan then moves on to part D, addressing the second derivative of f at x = -5 and x = 3. He calculates the second derivative at -5 as -1/2 and explains why it does not exist at x = 3 due to differing slopes from the right and left as x approaches 3. The segment concludes with a review of the scoring guidelines for question 3 and a teaser for the next video, which will cover question 4.
Mindmap
Keywords
π‘AP Calculus
π‘Free Response Questions
π‘Derivative
π‘Antiderivation
π‘Fundamental Theorem of Calculus
π‘Increasing Function
π‘Absolute Minimum
π‘Critical Numbers
π‘Second Derivative
π‘Semicircle and Line Segments
π‘Area Under the Curve
Highlights
Alan is teaching AP Calculus 2017 free response question 3
The question involves finding the values of F(-6) and F(5) using the graph of the derivative F'
F is the antiderivative of F', so we can find F(x) by integrating F' from -2 to x and adding F(-2)
F(-6) is calculated as 7 + (area from -2 to -6) = 7 - 4 = 3
F(5) is calculated as 7 + (area from -2 to 5) = 7 - 2pi + 3 = 10 - 2pi
To find where F is increasing, look where the derivative F' is above the x-axis between -6 and -2 and 2 and 5
To find the absolute minimum value of F, evaluate F at the critical points and endpoints: F(-6)=3, F(-2)=7, F(2)=7-2pi, F(5)=10-2pi
The absolute minimum value of F is 3 at x = -6
The second derivative F'' exists at x = -5 with value -1/2, but does not exist at x = 3 due to differing left/right limits
The scoring guidelines awarded full points for correctly using the initial condition, evaluating intervals, finding the minimum, and determining the second derivative
Alan provides a step-by-step walkthrough of solving the problem using the fundamental theorem of calculus
The area under the curve is calculated by integrating the derivative from the left endpoint to x, then adding the value of the function at the left endpoint
The function is increasing where the derivative is positive, which is determined by the graph of the derivative above the x-axis
The absolute minimum value is found by evaluating the function at critical points and endpoints, then comparing to determine the smallest value
The second derivative can be found by taking the derivative of the graph of the first derivative
The second derivative may not exist at certain points where the left and right limits do not agree
Alan's video provides a clear, detailed explanation of solving the AP Calculus problem using the fundamental theorem and other calculus concepts
The video is part of a series covering AP Calculus free response questions
Transcripts
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