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Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d²r/dt² = 2/t³; dr/dt|ₜ ₌ ₁ =1, r(1) = 1
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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 4.4.4
Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
4. y=9/14x^(1/3)(x^2-7)
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To find the inflection points, local maxima, and minima, start by taking the first derivative of the function y = \(\frac{9}{14}\)x^{\(\frac{1}{3}\)}(x^2 - 7). Use the product rule: if y = u*v, then y' = u'v + uv'.
Calculate the first derivative: Let u = \(\frac{9}{14}\)x^{\(\frac{1}{3}\)} and v = (x^2 - 7). Find u' and v'. u' = \(\frac{3}{14}\)x^{-\(\frac{2}{3}\)} and v' = 2x.
Substitute u, u', v, and v' into the product rule formula to find y': y' = \(\frac{3}{14}\)x^{-\(\frac{2}{3}\)}(x^2 - 7) + \(\frac{9}{14}\)x^{\(\frac{1}{3}\)}(2x).
To find critical points, set y' = 0 and solve for x. These points will help identify local maxima and minima. Analyze the sign changes of y' around these points to determine the nature of each critical point.
To find inflection points, take the second derivative of y and set it equal to zero. Solve for x to find where the concavity changes. Analyze the intervals where the second derivative is positive (concave up) and negative (concave down).
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Inflection Points
Inflection points occur where the concavity of a function changes from concave up to concave down or vice versa. To find these points, we need to determine where the second derivative of the function equals zero or is undefined, and verify a change in concavity. These points are crucial for understanding the overall shape and behavior of the graph.
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04:50
Critical Points
Local Maxima and Minima
Local maxima and minima are points where a function reaches a highest or lowest value, respectively, within a certain interval. These can be found by setting the first derivative to zero and using the second derivative test to confirm the nature of the critical points. Identifying these points helps in understanding the peaks and troughs of the graph.
Video consigliato:
07:09The First Derivative Test: Finding Local Extrema
Concavity and Differentiability
Concavity describes the direction a graph curves, either upwards (concave up) or downwards (concave down). A function is differentiable on intervals where it is smooth and continuous, without sharp corners or cusps. Analyzing concavity and differentiability provides insights into the function's behavior and helps identify intervals of interest for further analysis.
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05:59Determining Concavity Given a Function
Pratica correlata
Domanda del libro di testo
Identifying Extrema
In Exercises 15–18:
a. Find the open intervals on which the function is increasing and those on which it is decreasing.
b. Identify the function’s local and absolute extreme values, if any, saying where they occur.
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Finding Indefinite Integrals
In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
∫(−2cost) dt
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Domanda del libro di testo
Applications
Stopping a motorcycle The State of Illinois Cycle Rider Safety Program requires motorcycle riders to be able to brake from 30 mph (44 ft/sec) to 0 in 45 ft. What constant deceleration does it take to do that?
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Theory and Examples
[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.
f(x) = |x − 2| + |x + 3|, −5 ≤ x ≤ 5
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Domanda del libro di testo
Theory and Examples
[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.
h(x) = |x + 2| − |x − 3|, −∞ < x < ∞
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