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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.4.4
Chapter 4, Problem 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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1
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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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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.
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The 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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Determining Concavity Given a Function
Related Practice
Textbook Question

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.

7. y=sin|x|, -2π≤x≤2π

Textbook Question

Checking the Mean Value Theorem


Which of the functions in Exercises 7–12 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers.


f(x) = x⁴ᐟ⁵, [0, 1]

Textbook Question

Find values of a and b such that the function


ƒ(𝓍) = (a𝓍 + b) / 𝓍² ―1)


has a local extreme value of 1 at 𝓍 = 3. Is this extreme value a local maximum or a local minimum? Give reasons for your answer.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 3x⁻²ᐟ³, y(−1) = −5

Textbook Question

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 < ∞

Textbook Question

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.


∫(1 + tan²θ)dθ (Hint:1 + tan²θ = sec²θ)