Please find the local minimum and local maximum when x= Example 5The graph of a function fis given in the following figure.y=f(x)लिंज1 2 3 40ⒸMake a rough sketch of an antiderivative F, given that F(0) = 2.SolutionWe are guided by the fact that the slope of y = F(x) is f(x). We start at the point (0, 2) and draw F as an initially decreasing ✔✔✔ function since f(x) is negative ✔✔✔ when 0 < x < 1.Notice that f(1) = f(3) = 0,so F has horizontal tangents when x = 1 and x = 3. For 1 < x <3, f(x) is positiveand so F is increasing.We see that F has a local minimum when x =x and a local maximum when x =x. For x > 3, f(x) is negative, and so F is decreasing on (3,00).Since f(x) → 02✔We use this information to sketch the graph of the antiderivative in the figure below.as x→ 00, the graph of F becomes flatter as x→ ∞0. Also notice that F"(x) = f'(x) changes from positive to negative at x = 2 and from negative to positive at x = 41-y=F(x), so F has inflection points when x = 2 and x = 4

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Example 5 The graph of a function fis given in the following figure. f y=f(x) Q Make a rough sketch of an antiderivative F, given that F(0) = 2. Solution We are guided by the fact that the slope of y= F(x) is f(x). We start at the point (0, 2) and draw F as an initially decreasing ✔✔✔ function since f(x) is negative ✔✔✔ when 0 3, f(x) is negative, and so F is decreasing on (3,00). Since f(x) - 0 0 r 1 as x→ ∞o, the graph of F becomes flatter as x00. Also notice that F(x)= f'(x) changes from positive to negative at x = 2 and from negative to positive at x = 4 ✓ We use this information to sketch the graph of the antiderivative in the figure below. y= F(x) ✔ so F has inflection points when x = 2 and x = 4

Example 5 The graph of a function fis given in the following figure. f y=f(x) Q Make a rough sketch of an antiderivative F, given that F(0) = 2. Solution We are guided by the fact that the slope of y= F(x) is f(x). We start at the point (0, 2) and draw F as an initially decreasing ✔✔✔ function since f(x) is negative ✔✔✔ when 0 3, f(x) is negative, and so F is decreasing on (3,00). Since f(x) - 0 0 r 1 as x→ ∞o, the graph of F becomes flatter as x00. Also notice that F(x)= f'(x) changes from positive to negative at x = 2 and from negative to positive at x = 4 ✓ We use this information to sketch the graph of the antiderivative in the figure below. y= F(x) ✔ so F has inflection points when x = 2 and x = 4

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter3: Functions And Graphs

Section3.5: Graphs Of Functions

Problem 36E

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