Discuss the continuity of the function on the closed interval. Function Interval 2- x, F(x) = 1 2+-X, [-4, 6] x >0 O The function is discontinuous because f(-4) # f(6). O The function is continuous because the domain boundary is not inside the interval [-4, 6]. The function is discontinuous because all piecewise functions are discontinuous at their domain boundaries. The function is continuous because lim f(x) = lim f(x) = f(0) = 2. %3D %3D x-0 The function is continuous because lim f(x) and lim f(x) both exist. X→-4+ X-6

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Discuss the continuity of the function on the closed interval.
Function
Interval
2 x,
F(x) =
[-4, 6]
1
2+-X,
2
x > 0
O The function is discontinuous because f(-4) # f(6).
O The function is continuous because the domain boundary is not inside the interval [-4, 6].
The function is discontinuous because all piecewise functions are discontinuous at their domain boundaries.
The function is continuous because lim f(x) = lim f(x) = f(0) = 2.
x-0
O The function is continuous because lim f(x) and lim f(x) both exist.
X→-4+
X-6
Transcribed Image Text:Discuss the continuity of the function on the closed interval. Function Interval 2 x, F(x) = [-4, 6] 1 2+-X, 2 x > 0 O The function is discontinuous because f(-4) # f(6). O The function is continuous because the domain boundary is not inside the interval [-4, 6]. The function is discontinuous because all piecewise functions are discontinuous at their domain boundaries. The function is continuous because lim f(x) = lim f(x) = f(0) = 2. x-0 O The function is continuous because lim f(x) and lim f(x) both exist. X→-4+ X-6
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