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1.3 – Continuity, End Behavior, and Limits
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Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity test. If discontinuous, identify the type of discontinuity as infinite, jump, or removable. a. f(x) = 3x – 2 if x > -3; at x = -3 2 – x if x < - 3
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Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity test. If discontinuous, identify the type of discontinuity as infinite, jump, or removable. a. f(x) = 3x – 2 if x > -3; at x = -3 2 – x if x < - 3 1. Find f(-3).
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Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity test. If discontinuous, identify the type of discontinuity as infinite, jump, or removable. a. f(x) = 3x – 2 if x > -3; at x = -3 2 – x if x < - 3 1. Find f(-3). f(-3) = 2 – (-3) = 5
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Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity test. If discontinuous, identify the type of discontinuity as infinite, jump, or removable. a. f(x) = 3x – 2 if x > -3; at x = -3 2 – x if x < - 3 1. Find f(-3). f(-3) = 2 – (-3) = 5, so f(-3) exists
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2. Investigate values close to f(-3)
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2. Investigate values close to f(-3) x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)5.15.015.001-10.997-10.97-10.7
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2. Investigate values close to f(-3) As x -3 from left, f(x) 5 x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)5.15.015.001-10.997-10.97-10.7
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2. Investigate values close to f(-3) As x -3 from left, f(x) 5 As x -3 from right, f(x) -11 x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)5.15.015.001-10.997-10.97-10.7
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2. Investigate values close to f(-3) As x -3 from left, f(x) 5 As x -3 from right, f(x) -11 Since don’t approach same value, discontinuous and jump discontinuity. x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)5.15.015.001-10.997-10.97-10.7
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b. f(x) = x + 3; at x = -3 and x = 3 x 2 – 9
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b. f(x) = x + 3; at x = -3 and x = 3 x 2 – 9 1. Find f(-3) and f(3).
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b. f(x) = x + 3; at x = -3 and x = 3 x 2 – 9 1. Find f(-3) and f(3). f(-3) = -3 + 3 = 0 = Ø (-3) 2 – 9 0
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b. f(x) = x + 3; at x = -3 and x = 3 x 2 – 9 1. Find f(-3) and f(3). f(-3) = -3 + 3 = 0 = Ø (-3) 2 – 9 0 f(3) = 3 + 3 = 6 = Ø (3) 2 – 9 0
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b. f(x) = x + 3; at x = -3 and x = 3 x 2 – 9 1. Find f(-3) and f(3). f(-3) = -3 + 3 = 0 = Ø (-3) 2 – 9 0 f(3) = 3 + 3 = 6 = Ø (3) 2 – 9 0 Since both f(-3) = Ø and f(3) = Ø, f(x) is discontinuous at both x = -3 and x = 3.
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2. Investigate values close to f(-3) and f(3).
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x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169
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2. Investigate values close to f(-3) and f(3). As x -3 from left, f(x) -0.167 x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169
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2. Investigate values close to f(-3) and f(3). As x -3 from left, f(x) -0.167 As x -3 from right, f(x) -0.167 x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169
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2. Investigate values close to f(-3) and f(3). As x -3 from left, f(x) -0.167 As x -3 from right, f(x) -0.167 Since they approach same value, limit exists. x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169
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2. Investigate values close to f(-3) and f(3). As x -3 from left, f(x) -0.167 As x -3 from right, f(x) -0.167 Since they approach same value, limit exists. x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169 x2.92.992.9993.03.0013.013.1 f(x)f(x)-10-100-1000100010010
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2. Investigate values close to f(-3) and f(3). As x -3 from left, f(x) -0.167 As x -3 from right, f(x) -0.167 Since they approach same value, limit exists. As x 3 from left, f(x) -∞ x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169 x2.92.992.9993.03.0013.013.1 f(x)f(x)-10-100-1000100010010
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2. Investigate values close to f(-3) and f(3). As x -3 from left, f(x) -0.167 As x -3 from right, f(x) -0.167 Since they approach same value, limit exists. As x 3 from left, f(x) -∞ As x 3 from right, f(x) ∞ x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169 x2.92.992.9993.03.0013.013.1 f(x)f(x)-10-100-1000100010010
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2. Investigate values close to f(-3) and f(3). As x -3 from left, f(x) -0.167 As x -3 from right, f(x) -0.167 Since they approach same value, limit exists. As x 3 from left, f(x) -∞ As x 3 from right, f(x) ∞ Since limit x -3 exists but f(-3) doesn’t, removable discontinuity. x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169 x2.92.992.9993.03.0013.013.1 f(x)f(x)-10-100-1000100010010
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2. Investigate values close to f(-3) and f(3). As x -3 from left, f(x) -0.167 As x -3 from right, f(x) -0.167 Since they approach same value, limit exists. As x 3 from left, f(x) -∞ As x 3 from right, f(x) ∞ Since limit x -3 exists but f(-3) doesn’t, removable discontinuity. Since limit x -3 doesn’t exist, infinite discontinuity. x-3.1-3.01-3.001-3-2.999-2.99-2.9 f(x)f(x)-0.164-0.166-0.167 -0.169 x2.92.992.9993.03.0013.013.1 f(x)f(x)-10-100-1000100010010
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Ex. 2 Use the graph of the function to describe its end behavior.
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Ex. 2 Use the graph of the function to describe its end behavior. lim f(x) = - ∞ x - ∞
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Ex. 2 Use the graph of the function to describe its end behavior. lim f(x) = - ∞ x - ∞ lim f(x) = - ∞ x - ∞
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