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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
4장, 문제 4.1.27

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.


ƒ(x) = 3x³ + 3x² / 2 - 2x

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To find the critical points of the function \( f(x) = 3x^3 + \frac{3}{2}x^2 - 2x \), we first need to find its derivative, \( f'(x) \).
Differentiate the function: \( f'(x) = \frac{d}{dx}(3x^3) + \frac{d}{dx}(\frac{3}{2}x^2) - \frac{d}{dx}(2x) \).
Calculate each derivative: \( \frac{d}{dx}(3x^3) = 9x^2 \), \( \frac{d}{dx}(\frac{3}{2}x^2) = 3x \), and \( \frac{d}{dx}(2x) = 2 \).
Combine the derivatives to get \( f'(x) = 9x^2 + 3x - 2 \).
Set the derivative equal to zero to find the critical points: \( 9x^2 + 3x - 2 = 0 \). Solve this quadratic equation for \( x \) to find the critical points.

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Critical Points

Critical points of a function occur where its derivative is either zero or undefined. These points are essential for identifying local maxima, minima, and points of inflection. To find critical points, one typically takes the derivative of the function and solves for the values of x that satisfy the condition.
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04:50
Critical Points

Derivative

The derivative of a function measures the rate at which the function's value changes as its input changes. It is a fundamental concept in calculus, representing the slope of the tangent line to the curve at any given point. For polynomial functions, the derivative can be calculated using power rules, which simplify the process of finding critical points.
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Polynomial Functions

Polynomial functions are expressions that involve variables raised to whole number powers, combined using addition, subtraction, and multiplication. They are characterized by their degree, which is the highest power of the variable. Understanding the behavior of polynomial functions, including their critical points and end behavior, is crucial for analyzing their graphs and determining local extrema.
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6:04
Introduction to Polynomial Functions
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x⁴ - x² + y² = 0 (Figure-8 curve)

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ƒ(x) = x/(x²+9)⁵ on [-2,2]

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f(x) = x² - 10; x₀ = 3

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f(x) = x ln (x + 1) -1 ; x₀ = 1.7

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ƒ(x) = (4x³/3) + 5x² - 6x on [0,5]

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f(x) = eˣ/(e²ᵉ + 1)

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