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
and the answer options are
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
To describe the continuity of h(x) at x = 0, we need to examine the behavior of both f(x) and g(x) around x = 0.
First, let's look at the graph of f(x). We see that f(x) has a discontinuity at x = 0 since there is a jump in the graph. However, there is also a point at (0,0) on the graph of f(x) which makes a parabola shape. This means that for values of x close to 0, the graph of f(x) becomes continuous again.
Next, let's consider the graph of g(x). We see that g(x) has a point at (0,-1), which indicates that g(x) is continuous at x = 0.
Now, since h(x) is defined as the composition of f(g(x)), we can evaluate the continuity of h(x) at x = 0 by considering the behavior of f(g(x)) around x = 0.
When x is close to 0, g(x) approaches the value -1, and as we discussed earlier, f(x) becomes continuous again when x is close to 0. Therefore, as x approaches 0, g(x) approaches -1 and f(g(x)) approaches f(-1) (which is a continuous point on the graph of f(x)).
Based on this analysis, we can conclude that h(x) is continuous at x = 0 because both f(g(x)) and g(x) are continuous at x = 0.