2011 Calculus AB free response #6b | AP Calculus AB | Khan Academy
TLDRThe video script discusses the concept of a piecewise-defined function and its derivative, specifically focusing on a function where the derivative is not defined at x=0 due to different behaviors from the left and right approaches. The script then methodically calculates the derivative for other values of x, resulting in a piecewise function. It proceeds to solve for x when the derivative equals -3, concluding that the solution must lie in the part of the function where x is greater than 0. Through algebraic manipulation and the application of natural logarithms, the script finds that x equals -1/4 times the natural log of 3/4, confirming this solution is valid for the given domain.
Takeaways
- π The problem involves finding the derivative of a function, f'(x), as a piecewise-defined function, with x β 0.
- π€ The derivative is not defined at x = 0 because the behavior of the derivative changes when approaching 0 from the left versus the right.
- π For x < 0, the derivative is defined as -2cos(x), since the derivative of sin(x) is cos(x) and the function is multiplied by -2.
- π For x > 0, the derivative is defined as -4e^(-4x), using the chain rule to differentiate -4x and e^(-4x).
- π’ The derivative is undefined at x = 0, which is explicitly stated in the problem setup.
- π To find the value of x where f'(x) = -3, we need to examine the piecewise-defined derivative to determine which part of the function can achieve this value.
- π« The function -2cos(x) cannot equal -3 because the cosine function is bounded between -1 and 1.
- π§ We deduce that the part of the derivative for x > 0, -4e^(-4x), is the one that could potentially equal -3.
- π By setting -4e^(-4x) equal to -3 and solving for x, we find that x = -1/4 * ln(3/4), ensuring that this value of x is greater than 0.
- π The final answer is x = -1/4 * ln(3/4), where the derivative of the function equals -3.
Q & A
Why is the derivative not defined at x=0 in the given function?
-The derivative is not defined at x=0 because the behavior of the derivative changes when approaching 0 from the left versus from the right, leading to different expressions.
What is the piecewise definition of f prime of x for the given function?
-For x < 0, f prime of x is -2cos(x), and for x > 0, f prime of x is -4e^(-4x).
How do we determine the value of x for which f prime of x equals negative 3?
-We need to consider the piecewise definition and find the value of x that satisfies the equation for the part of the function that can take on values less than or equal to -3.
Why can't the value of x that makes f prime of x equal negative 3 be less than 0?
-The value of x cannot be less than 0 because -2cos(x) is bounded between -2 and 2, and thus can never reach -3.
What is the significance of the bounded nature of the cosine function in this context?
-The bounded nature of the cosine function indicates that -2cos(x) cannot be equal to -3, so the solution must lie in the part of the function where x is greater than 0.
How do we solve for x when f prime of x equals negative 3?
-We set the expression for the derivative when x > 0 equal to -3, solve for e^(-4x), and then take the natural logarithm of both sides to find the value of x.
What is the mathematical expression for the value of x that makes f prime of x equal negative 3?
-x = -1/4 * ln(3/4), where ln represents the natural logarithm.
How do we verify that the found value of x is greater than 0?
-We recognize that the natural log of 3/4 is negative and when multiplied by a negative (-1/4), the result is positive, thus ensuring that x is greater than 0.
What is the final expression for f prime of x when x equals the found value?
-f prime of (-1/4 * ln(3/4)) equals -3, confirming that our solution is correct.
What is the practical implication of finding the derivative not defined at a point?
-It indicates a discontinuity or a sharp turn in the function at that point, which can be crucial for understanding the function's behavior and for applications in physics or engineering where smoothness or differentiability is often required.
Why is it important to consider the domain of the trigonometric and exponential functions when solving this problem?
-Considering the domain and behavior of these functions helps us to understand the range of possible values for the derivative, which is essential for narrowing down the correct piece of the piecewise function to use in solving the problem.
Outlines
π Piecewise Derivative and Critical Value Analysis
This paragraph discusses the process of expressing the derivative of a function, denoted as f prime of x, as a piecewise-defined function, particularly when x is not equal to 0. It delves into the rationale behind excluding x=0 from the domain of the derivative, highlighting the difference in the derivative's behavior when approaching x=0 from the left versus the right. The paragraph then proceeds to calculate the derivative for all other values of x, detailing the application of basic derivative rules for functions like sine and exponentials. A key focus is on finding the specific value of x for which the derivative equals negative 3. By analyzing the bounded nature of the cosine function, the paragraph eliminates the possibility of the negative 2 cosine term reaching negative 3 and concludes that the solution must lie within the exponential part of the function. Through algebraic manipulation and the application of natural logarithms, the paragraph arrives at the solution x = -1/4 * natural log(3/4), verifying that this value is indeed greater than 0, as required by the chosen piecewise function segment.
π Final Derivative Expression and Solution Verification
In this paragraph, the final step of the mathematical analysis is presented, where the previously found value of x is used to express the derivative function f prime of x as a product of a constant and the natural logarithm function. The paragraph reiterates the solution as f prime of (-1/4 * natural log(3/4)) equals negative 3, thereby completing the task of finding the critical value for which the derivative meets a specified condition. This concise summary encapsulates the conclusion of the piecewise derivative analysis and its successful application in solving for the critical value.
Mindmap
Keywords
π‘Derivative
π‘Piecewise-defined function
π‘Discontinuity
π‘Cosine function
π‘Chain rule
π‘Natural logarithm
π‘Exponential function
π‘Slope
π‘Algebra
π‘Bounded function
π‘Positive and negative values
Highlights
The derivative is not defined at x=0 due to different behaviors when approaching from the left versus the right.
For x < 0, the derivative is -2cos(x), as derived from the first part of the function.
For x > 0, the derivative is -4e^(-4x), obtained using the chain rule on the second part of the function.
The piecewise definition of the derivative is crucial for solving the problem, as different cases must be considered separately.
Cosine is a bounded function, ranging between -1 and 1, which helps in eliminating the first case for the value of -3.
The equation -4e^(-4x) = -3 is set up to find the value of x for which the derivative equals -3.
By dividing both sides by -4, the equation simplifies to e^(-4x) = 3/4.
Taking the natural log of both sides results in -4x = ln(3/4), allowing for the solution of x.
Solving for x gives us x = -1/4 * ln(3/4), which is the value for which the derivative is -3.
It is verified that the value of x is greater than 0, which validates the use of the second case of the derivative definition.
The natural log of 3/4 is negative, but the overall value of x is positive due to the negative signs involved.
The final answer is x = -1/4 * ln(3/4), which corresponds to the derivative being -3.
The process demonstrates the importance of understanding the domain and behavior of functions when finding derivatives.
The use of piecewise functions allows for the handling of discontinuities and different behaviors in mathematical functions.
The chain rule is a fundamental tool in calculus for finding derivatives of composite functions.
This problem showcases the application of derivative concepts in solving for specific function values.
The approach to solving the problem involves a combination of algebraic manipulation and understanding of function properties.
The solution requires a methodical process of elimination and the correct application of mathematical rules.
The problem highlights the practical use of derivatives in analyzing the rate of change of functions at specific points.
Transcripts
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