Textbook Question
Parallel tangent lines Find the two points where the curve x² + xy + y² = 7 crosses the x-axis, and show that the tangent lines to the curve at these points are parallel. What is the common slope of these tangent lines?
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.2.39
One-Sided Derivatives
Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.
Verified step by step guidance1
Identify the function and the point of interest: The function is piecewise, with \( y = \sqrt{x} \) for \( x < 1 \) and \( y = 2x - 1 \) for \( x \geq 1 \). The point of interest is \( P(1, 1) \).
Compute the left-hand derivative at \( x = 1 \): Use the definition of the derivative as a limit. For \( x < 1 \), the function is \( y = \sqrt{x} \). The left-hand derivative is \( \lim_{{h \to 0^-}} \frac{\sqrt{1+h} - 1}{h} \).
Compute the right-hand derivative at \( x = 1 \): For \( x \geq 1 \), the function is \( y = 2x - 1 \). The right-hand derivative is \( \lim_{{h \to 0^+}} \frac{(2(1+h) - 1) - 1}{h} \).
Evaluate the left-hand limit: Simplify the expression \( \frac{\sqrt{1+h} - 1}{h} \) using algebraic manipulation, such as multiplying by the conjugate, to find the limit as \( h \to 0^- \).
Evaluate the right-hand limit: Simplify the expression \( \frac{2h}{h} \) to find the limit as \( h \to 0^+ \). Compare the left-hand and right-hand derivatives to determine if they are equal.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
One-Sided Derivatives
One-sided derivatives are limits that evaluate the behavior of a function as it approaches a specific point from one side only. The right-hand derivative considers the limit as the input approaches the point from the right, while the left-hand derivative considers the limit from the left. If these two limits exist but are not equal, the function is not differentiable at that point.
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One-Sided Limits
Differentiability
A function is differentiable at a point if it has a defined derivative at that point, which means the function must be continuous and have a unique tangent line. If the left-hand and right-hand derivatives at a point are not equal, the function is not differentiable there. This concept is crucial for understanding the smoothness and behavior of functions at specific points.
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Limits
Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. They are essential for defining derivatives, as the derivative itself is the limit of the average rate of change of the function as the interval shrinks to zero. Understanding limits allows for the analysis of function behavior near points of interest, particularly in determining continuity and differentiability.
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Related Practice
Textbook Question
In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.
__
x + √xy = 6, (4, 1)
Textbook Question
[Technology Exercise]
Graph the curves in Exercises 39–48.
a. Where do the graphs appear to have vertical tangent lines?
b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38.
y = 4x²/⁵ − 2x
Textbook Question
Suppose that the function v in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule?
Textbook Question
In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?
54. x³ + y² = sin²y
Textbook Question
Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is,
d/dx (x⁻ᵐ) = −mx⁻ᵐ⁻¹
where m is a positive integer.