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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 2 - Limits and ContinuityProblem 2.1.18
Chapter 2, Problem 2.1.18

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=√7−x, P(−2,3)

Verified step by step guidance
1
First, identify the function given: \( y = \sqrt{7 - x} \). We need to find the derivative of this function to determine the slope of the curve at the point P.
To find the derivative, use the chain rule. The outer function is \( \sqrt{u} \) and the inner function is \( u = 7 - x \). The derivative of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \) and the derivative of \( 7 - x \) is \( -1 \).
Apply the chain rule: \( \frac{dy}{dx} = \frac{1}{2\sqrt{7 - x}} \cdot (-1) = -\frac{1}{2\sqrt{7 - x}} \). This is the derivative of the function, which gives the slope of the tangent line at any point on the curve.
Substitute the x-coordinate of point P, which is \( x = -2 \), into the derivative to find the slope at P: \( m = -\frac{1}{2\sqrt{7 - (-2)}} = -\frac{1}{2\sqrt{9}} = -\frac{1}{6} \).
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 \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point P(-2, 3). Substitute \( m = -\frac{1}{6} \), \( x_1 = -2 \), and \( y_1 = 3 \) into the equation to get the equation of the tangent line.

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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 provides the slope of the tangent line to the curve at that point. It is a fundamental concept in calculus that measures how a function changes as its input changes. For the function y = √(7−x), finding the derivative will help determine the slope at the point P(−2,3).
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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 equation of the tangent line can be found using the slope from the derivative and the coordinates of the point. For the curve y = √(7−x) at P(−2,3), the tangent line represents the best linear approximation of the curve near this point.
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Slopes of Tangent Lines

Point-Slope Form

The point-slope form of a line's equation is useful for writing the equation of a tangent line. It is expressed as y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line. Using the slope from the derivative and the point P(−2,3), this form helps in constructing the equation of the tangent line to the curve.
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Related Practice
Textbook Question

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Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form limh→0 (f(x+h) − f(x)) / h occur frequently in calculus. In Exercises 57–62, evaluate this limit for the given value of x and function f.


f(x) = 3x - 4, x = 2

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Find the limits in Exercises 37–48. Write ∞ or −∞ where appropriate.


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If 2−x² ≤ g(x) ≤ 2cosx for all x, find limx→0 g(x).

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Limits with trigonometric functions


Find the limits in Exercises 43–50.


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Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


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Textbook Question

Use formal definitions to prove the limit statements in Exercises 93–96.


lim x → 0 (−1 / x²) = −∞