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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 2 - Limits and ContinuityProblema 2.1.18
Capitolo 2, Problema 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)

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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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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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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 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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Percorso guidato
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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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Slope-Intercept Form
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Limits of Average Rates of Change


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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Average Rates of Change


In Exercises 1–6, find the average rate of change of the function over the given interval or intervals.


P(θ)=θ³ − 4θ² + 5θ; [1,2]

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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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Limits with trigonometric functions


Find the limits in Exercises 43–50.


lim x→0 tan x

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Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limθ→0 sin θ / sin 2θ

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Finding Deltas Algebraically


Each of Exercises 15–30 gives a function f(x) and numbers L, c, and ε>0. In each case, find the largest open interval about c on which the inequality |f(x)−L| <ε holds. Then give a value for δ>0 such that for all x satisfying 0 < |x−c| < δ, the inequality |f(x)−L| < ε holds.


f(x) = 1/x, L = 1/4, c = 4, ε = 0.05

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