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Ch. 4 - Applications of Derivatives
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. 4 - Applications of DerivativesProblema 4.3.44a
Capitolo 4, Problema 4.3.44a

Identifying Extrema


In Exercises 41–52:


a. Identify the function’s local extreme values in the given domain, and say where they occur.


g(x) = −x² − 6x − 9,−4 ≤ x < ∞

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To find the local extrema of the function \( g(x) = -x^2 - 6x - 9 \), we first need to find its derivative. The derivative \( g'(x) \) will help us identify critical points where the slope of the tangent is zero or undefined.
Calculate the derivative of \( g(x) \). Using the power rule, the derivative is \( g'(x) = -2x - 6 \).
Set the derivative equal to zero to find the critical points: \( -2x - 6 = 0 \). Solve for \( x \) to find the critical point.
Once the critical point is found, determine whether it is a local maximum or minimum by using the second derivative test. Calculate the second derivative \( g''(x) \) and evaluate it at the critical point.
Finally, check the endpoints of the domain \( -4 \leq x < \infty \) to ensure there are no other local extrema. Since the domain is unbounded on the right, focus on the behavior of the function as \( x \to -4 \) and as \( x \to \, \infty \).

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Local Extrema

Local extrema refer to the points in a function where it reaches a local maximum or minimum within a specific interval. These are points where the function changes direction, and can be identified by finding where the derivative equals zero or is undefined. Understanding local extrema is crucial for analyzing the behavior of functions within a given domain.
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Finding Extrema Graphically

Derivative

The derivative of a function represents the rate of change of the function's value with respect to its input. It is a fundamental tool in calculus for finding slopes of tangent lines and identifying critical points, which are potential locations for local extrema. Calculating the derivative of g(x) helps determine where the function's slope is zero, indicating possible extrema.
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Derivatives

Quadratic Functions

Quadratic functions are polynomial functions of degree two, typically in the form ax² + bx + c. They graph as parabolas, which can open upwards or downwards. The vertex of the parabola represents the function's maximum or minimum value, depending on the direction it opens. For g(x) = −x² − 6x − 9, understanding its quadratic nature helps identify the vertex as the point of local extremum.
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Introduction to Polynomial Functions
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Domanda del libro di testo

53. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day.

a. Start counting time with t=0 at noon and express the distance s between the ships as a function of t.

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Finding displacement from an antiderivative of velocity


a. Suppose that the velocity of a body moving along the s-axis is


ds/dt = v = 9.8t − 3.


i. Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0.

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Checking Antiderivative Formulas


Right, or wrong? Say which for each formula and give a brief reason for each answer.


∫tanθ sec²θ dθ = sec³θ / 3 + C

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Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

πcos πx

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Identifying Extrema


In Exercises 41–52:


a. Identify the function’s local extreme values in the given domain, and say where they occur.


g(x) = x² − 4x + 4, 1 ≤ x < ∞

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Analyzing Functions from Derivatives


Answer the following questions about the functions whose derivatives are given in Exercises 1–14:


a. What are the critical points of f?


f′(x) = (x − 1)(x + 2)(x − 3)

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