The graph of a function f is given.   The x y-coordinate plane is given. A curve with 3 parts is graphed. The first part is linear, enters the window in the second quadrant, goes down and right, crosses the x-axis at approximately x = −0.33, crosses the y-axis at y = −0.25, and ends at the open point (1, −1). The second part is the point (1, 1). The third part is linear, begins at the open point (1, −1), goes up and right, crosses the x-axis at x = 2, and exits the window in the first quadrant. Determine whether f is continuous on its domain. continuous not continuous     If it is not continuous on its domain, say why. lim  x→1+  f(x) ≠  lim  x→1−  f(x), so  lim  x→1  f(x) does not exist. The function is not defined at x = 1.     The graph is continuous on its domain. lim  x→1  f(x) = −1 ≠ f(1)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter3: Linear And Nonlinear Functions
Section3.1: Graphing Linear Functions
Problem 5CYU
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The graph of a function f is given.

 

The x y-coordinate plane is given. A curve with 3 parts is graphed.

  • The first part is linear, enters the window in the second quadrant, goes down and right, crosses the x-axis at approximately x = −0.33, crosses the y-axis at y = −0.25, and ends at the open point (1, −1).
  • The second part is the point (1, 1).
  • The third part is linear, begins at the open point (1, −1), goes up and right, crosses the x-axis at x = 2, and exits the window in the first quadrant.

Determine whether f is continuous on its domain.

continuous

not continuous    

If it is not continuous on its domain, say why.

lim 

x→1+ 

f(x) ≠ 

lim 

x→1− 

f(x), so 

lim 

x→1 

f(x) does not exist.

The function is not defined at x = 1.    

The graph is continuous on its domain.

lim 

x→1 

f(x) = −1 ≠ f(1)

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