University:
Kentucky State University
Course:
MAT 120 | Precalculus
Academic year:

2021
Views:

323

Pages:

8
Author:

agatoary6zt

Maximum & Minimum Value Dec 30, 2021 *Def* For a function f(x) defined on a set S of x's and c ∈ S * f(c) is absolute max for f(x) if: f(c) ≥ f(x) for all x ∈ S * f(c) is absolute min for f(x) if: f(c) ≤ f(x) for all x ∈ S *Def* f(c) is a local max for f(x) if: f(c) ≥ f(x) for all x in some open interval containing c. * f(c) is a local min for f(x) if: f(c) ≤ f(x) for all x in some open interval containing c. [Image of a graph with local maxima and minima] *Thrm: If f(x) is Continuous on [a,b], then f(x) attains Max & Min *Defn: A number c ∈ Domain f(x) is called a critical number of f(x) if: f'(c) = 0 or f'(c) DNE *Thrm: If f(c) is a local extremum "Max or Min" then c must be a critical number for f(x), but the converse is not true *Defn: A function f(x) is strictly increasing for all x ∈ open interval (I) then f(x₁) < f(x₂) for all x₁ < x₂ ∈ I * A function f(x) is strictly decreasing for all x ∈ open interval (I) then f(x₁) > f(x₂) for all x₁ < x₂ ∈ I * **Theorem:** Suppose f(x) is differentiable on I. Then: * If f'(x) > 0 for all x ∈ I, then f(x) is increasing on I. * If f'(x) < 0 for all x ∈ I, then f(x) is decreasing on I. * If f'(x) = 0 for all x ∈ I, then f(x) is constant on I. * **First Derivative Test:** * Let f(x) be continuous on [a,b] and differentiable on (a,b). * Let c ∈ (a,b) be a critical number. Then: * If f'(x) > 0 for all x ∈ (a,c) and f'(x) < 0 for all x ∈ (c,b), then (c, f(c)) is a local maximum. * If f'(x) < 0 for all x ∈ (a,c) and f'(x) > 0 for all x ∈ (c,b), then (c, f(c)) is a local minimum. * **Definition:** For a function f(x) that is differentiable on I, then: * f(x) is concave up on the interval I if: * f'(x) is increasing on I * f''(x) ≥ 0 for all x ∈ I * f(x) is concave down on the interval I if: * f'(x) is decreasing on I * f''(x) ≤ 0 for all x ∈ Definition: Suppose f(x) is cont on [a, b] and the graph changes concavity at point (c, f(c)), c ∈ (a, b). Then the point (c, f(c)) is an inflection point. eg) Let f(x) = x³ - 12x, [0,4] 1. Find the increasing & decreasing 2. Find the extreme point 3. Find the domains of concavity 4. Find such that f'(x) has critical point 5. Find the inflection point Domain: [0,4] f'(x) = 3x² - 12 = 3(x² - 4) = 3(x + 2)(x - 2) = 0 Critical points: x = 2, 0, 4 f'(x) = 3x² - 12 Increasing: (2, ∞) Decreasing: (0, 2) f(0) = 0³ - 12(0) = 0 Nothing (bc it's on a closed interval) f(4) = 4³ - 12(4) = 16 Abs. Max f''(x) = 6x - 12 = 6(x - 2) = 0 x = 2 ∈ Dom Concave up: (2, 4) There's no inflection point eg) f(x) = x³ - 6x² + 1 (complete all the above steps) 1. Dom = ℝ 2. f'(x) = 3x² - 12x = 0 3x(x - 4) = 0 x = 0 or x = 4 *Critical no. = {0, 4} / Critical points = (0, f(0)), (4, f(4)) * Increasing: (-∞, 0) U (4, ∞) * Decreasing: (0, 4) (0, f(0)) Local Max (4, f(4)) Local Min f''(x) = 3x² - 12 = 0 f''(x) = 6(x - 2) = 0 x = 2 * Concave down: (-∞, 2) * Concave up: (2, ∞) (2, f(2)) Inflection point