Suppose h(x) = f (g(x)) and the graphs of f and g are shown below. Describe the continuity of h at x = 0.
how would you do this if you were given two graphs:
discontinuity at -2,0; point at -1,1; and point at 0,0 (to make a parabola)
then discontinuity at 0,-1; and then continuous points 1,-1; 2,-1, 3,-1....\
G- point at-2,-1; -1,0; discontinuity at 0,1
then point at 2,0; 1,1; 2,0; 3,1; 4,2
h is continuous at x = 0.
h is discontinuous at x = 0 since h(0) is undefined.
h is discontinuous at x = 0 since the limit of h(x) as x approaches 0 does not exist
h is discontinuous at x = 0 since the limit of h(x) as x approaches 0 does not equal 0
Hard to tell which branch of the functions are defined at the discontinuity. That is, where is the solid dot, and where is the open circle?
g(0) is hard to figure, since "discontinuity at 0,1" doesn't give a clear picture. Also, you have (2,0) twice in the list of g's values.
To determine the continuity of h at x = 0 based on the given graphs of f and g, we need to consider two things: the values of f(g(x)) as x approaches 0 and the existence of h(0).
First, let's evaluate f(g(x)) as x approaches 0. Looking at the graph of f, we see that f(g(0)) would be undefined since g(0) is at a point of discontinuity for f. Therefore, we can conclude that h is discontinuous at x = 0 because h(0) is undefined.
Alternatively, we can also consider the limit of h(x) as x approaches 0. To find the limit, we need to evaluate f(g(x)) as x approaches 0 from both the left and the right sides. From the given information, we can see that the values of g(x) approach 0 as x approaches 0 from either side. However, the corresponding values of f(g(x)) do not approach a defined value or they do not approach each other. Therefore, the limit of h(x) as x approaches 0 does not exist. Hence, h is discontinuous at x = 0.
In summary, based on the given graphs, we can conclude that h is discontinuous at x = 0 because h(0) is undefined and the limit of h(x) as x approaches 0 does not exist.