Textbook Question
In Exercises 1–4, say whether the function graphed is continuous on [−1, 3]. If not, where does it fail to be continuous and why?
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Ch. 2 - Limits and Continuity
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 2, Problem 2.1.14
Slope of a Curve at a Point
In Exercises 7–18, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.
y=x³−3x²+4, P(2,0)
Verified step by step guidance1
First, find the derivative of the function y = x³ - 3x² + 4. The derivative, y', represents the slope of the curve at any point x. Use the power rule for differentiation: if y = xⁿ, then y' = n*xⁿ⁻¹.
Apply the power rule to each term in the function: the derivative of x³ is 3x², the derivative of -3x² is -6x, and the derivative of the constant 4 is 0. Therefore, the derivative of the function is y' = 3x² - 6x.
Evaluate the derivative at the given point P(2,0) to find the slope of the curve at this point. Substitute x = 2 into the derivative: y'(2) = 3(2)² - 6(2).
Now, use the point-slope form of the equation of a line to find the equation of the tangent line at P. The point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point P(2,0).
Substitute the slope found in step 3 and the coordinates of point P into the point-slope form to get the equation of the tangent line. This will give you the equation of the tangent line at the point P(2,0).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative
The derivative of a function at a point measures the rate at which the function's value changes as its input changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In the context of a curve, the derivative at a specific point gives the slope of the tangent line to the curve at that point.
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Derivatives
Tangent Line
A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. The equation of the tangent line can be expressed in point-slope form, which utilizes the slope and the coordinates of the point of tangency.
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05:13Slopes of Tangent Lines
Point-Slope Form
The point-slope form of a linear equation is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. This form is particularly useful for writing the equation of a tangent line once the slope has been determined. By substituting the slope and the coordinates of the point of tangency into this formula, one can easily derive the equation of the tangent line.
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Guided course
3:56Slope-Intercept Form
Related Practice
Textbook Question
Use formal definitions to prove the limit statements in Exercises 93–96.
lim x → 0 (−1 / x²) = −∞
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Textbook Question
Domains and Asymptotes
Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.
y = (√(x² + 4)) / x
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Textbook Question
Explain why the equation cos x = x has at least one solution.
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Textbook Question
Limits of Rational Functions
In Exercises 13–22, find the limit of each rational function (a) as x → ∞ and (b) as x → −∞. Write ∞ or −∞ where appropriate.
f(x) = (2x + 3)/(5x + 7)
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Textbook Question
Removable discontinuity Give an example of a function f (x) that is continuous for all values of x except x = 2, where it has a removable discontinuity. Explain how you know that f is discontinuous at x = 2, and how you know the discontinuity is removable.
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