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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Hass 15th EditionCh. 4 - Applications of Derivatives문제 4.4.119
4장, 문제 4.4.119

119. Find the values of constants a, b, and c such that the graph of y = ax^3 + bx^2 + cx has a
local maximum at x = 3, local minimum at x =- 1, and inflection point at (1, 11).

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To find the values of a, b, and c, start by using the conditions for local maxima and minima. The first derivative of the function y = ax^3 + bx^2 + cx is y' = 3ax^2 + 2bx + c. Set y'(3) = 0 for the local maximum at x = 3, and y'(-1) = 0 for the local minimum at x = -1.
Next, use the condition for the inflection point. The second derivative of the function is y'' = 6ax + 2b. Set y''(1) = 0 to find the inflection point at x = 1.
Substitute x = 3 into y' = 3ax^2 + 2bx + c to get the equation 27a + 6b + c = 0. This is the first equation.
Substitute x = -1 into y' = 3ax^2 + 2bx + c to get the equation 3a - 2b + c = 0. This is the second equation.
Substitute x = 1 into y'' = 6ax + 2b to get the equation 6a + 2b = 0. This is the third equation. Solve this system of equations to find the values of a, b, and c.

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주요 개념

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Critical Points and Derivatives

Critical points occur where the first derivative of a function is zero or undefined, indicating potential local maxima or minima. To find these points for the function y = ax^3 + bx^2 + cx, compute the first derivative and set it to zero. Solving this will help identify the x-values where the local maximum and minimum occur, which are given as x = 3 and x = -1.
추천 영상:
04:50
Critical Points

Second Derivative Test

The second derivative test helps determine the nature of critical points. If the second derivative at a critical point is positive, the point is a local minimum; if negative, it's a local maximum. For the function y = ax^3 + bx^2 + cx, evaluate the second derivative at x = 3 and x = -1 to confirm the local maximum and minimum, respectively.
추천 영상:
06:02
The Second Derivative Test: Finding Local Extrema

Inflection Points

An inflection point is where the function changes concavity, identified by setting the second derivative to zero and confirming a sign change. For y = ax^3 + bx^2 + cx, the inflection point is given at (1, 11). Ensure the second derivative equals zero at x = 1 and verify the function's value at this point matches the y-coordinate of the inflection point.
추천 영상:
04:50
Critical Points
관련 실천
교과서 질문

Root Finding

5. Use Newton's method to find the positive fourth root of 2 by solving the equation x^4 -2 = 0. Start with x_0 = 1 and find x_2.

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교과서 질문

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

71. y' = x(x² - 12)

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교과서 질문

Finding Extrema from Graphs


In Exercises 1–6, determine from the graph whether the function has any absolute extreme values on [a, b]. Then explain how your answer is consistent with Theorem 1.


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교과서 질문

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

74. y' = (x² - 2x)(x - 5)²

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교과서 질문

Finding Extreme Values

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.


y = 𝓍³ + 𝓍² ― 8𝓍 + 5

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교과서 질문

Sketch the graphs of the rational functions in Exercises 53–60.


𝓍⁴ ― 1

y = ------------------

𝓍²

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