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Theory and Examples
In Exercises 51 and 52, give reasons for your answers.
Let f(x) = (x − 2)²ᐟ³.
a. Does f′(2) exist?
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- 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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 4, Problema 4.3.57a
Identifying Extrema
In Exercises 53–60:
a. Find the local extrema of each function on the given interval, and say where they occur.
f(x) = x / 2 − 2sin (x/2), 0 ≤ x ≤ 2π
Guida verificata passo dopo passo1
To find the local extrema of the function \( f(x) = \frac{x}{2} - 2\sin\left(\frac{x}{2}\right) \) on the interval \( 0 \leq x \leq 2\pi \), we first need to find the derivative \( f'(x) \).
Differentiate the function: \( f'(x) = \frac{1}{2} - \cos\left(\frac{x}{2}\right) \). This derivative will help us find the critical points where the slope of the tangent is zero or undefined.
Set the derivative equal to zero to find critical points: \( \frac{1}{2} - \cos\left(\frac{x}{2}\right) = 0 \). Solve for \( x \) to find the critical points within the interval.
Evaluate the function \( f(x) \) at the critical points and at the endpoints of the interval \( x = 0 \) and \( x = 2\pi \) to determine the local extrema.
Compare the values of \( f(x) \) at these points to identify the local maximum and minimum values, and specify where they occur within the interval.
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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 given interval. To find these points, one typically examines the derivative of the function to identify critical points where the derivative is zero or undefined, and then uses the second derivative test or evaluates the function at these points to determine the nature of the extrema.
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05:58
Finding Extrema Graphically
Derivative and Critical Points
The derivative of a function provides the rate of change of the function with respect to its variable. Critical points occur where the derivative is zero or undefined, indicating potential locations for local extrema. By solving f'(x) = 0, we can find these critical points, which are essential for determining where the function's slope changes direction, potentially indicating maxima or minima.
Video consigliato:
04:50Critical Points
Interval Analysis
Interval analysis involves examining the behavior of a function within a specified range of values. For the function f(x) = x/2 − 2sin(x/2) on the interval [0, 2π], it is crucial to evaluate the function at the endpoints and any critical points within this interval to identify local extrema. This ensures that all potential maxima and minima are considered within the given domain.
Video consigliato:
02:59Finding Area Between Curves that Cross on the Interval
Pratica correlata
Domanda del libro di testo
Theory and Examples
In Exercises 51 and 52, give reasons for your answers.
Let f(x) = |x³ − 9x|.
a. Does f'(0) exist?
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Domanda del libro di testo
52. Two masses hanging side by side from springs have positions s_1 = 2 sin t and s_2 = sin 2t,
respectively.
a. At what times in the interval 0 < t do the masses pass each other? (Hint: sin 2t = 2 sint cost.)
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Domanda del libro di testo
Identifying Extrema
In Exercises 53–60:
a. Find the local extrema of each function on the given interval, and say where they occur.
f(x) = sin 2x, 0 ≤ x ≤ π
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Domanda del libro di testo
Identifying Extrema
In Exercises 41–52:
a. Identify the function’s local extreme values in the given domain, and say where they occur.
k(x) = x³ + 3x² + 3x + 1, −∞ < x ≤ 0
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Domanda del libro di testo
Finding Antiderivatives
In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
1 / x²
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