Below is the graph of y = g'(x), (which looks like the following):
f(x) { 3 x<2}
x-3 x>2
Here are the questions:
Answer the following questions about the function g(x) .
A. Is is possible, impossible, or certain that g is continuous at x=2? Explain.
B. Is it possible, impossible, or certain that g(2)=0? Explain.
C. Is it possible, impossible, or certain that g has a vertical asymptote at x = 2? Explain.
D. Is it possible, impossible, or certain that g has a jump discontinuity at x = 2?
E. Is it possible, impossible, or certain that g has a removable discontinuity at x = 2?
My answers are the following:
A) impossible
B) impossible
C) ipossible
D)possible
E) certain
looks good
The link to the graph is:
www.desmos.com/calculator/mrmzffnwao
All my answers are correct?
To answer these questions, let's first understand what the graph of y = g'(x) tells us about the function g(x).
The graph of y = g'(x) gives us the derivative of the function g(x). The derivative represents the rate of change of the function at different points. In this case, the graph consists of two linear segments:
1. For x < 2, the slope of the function is 3. This means that for values of x less than 2, the function g(x) is a constant function with a value of 3.
2. For x > 2, the slope of the function is 1. This means that for values of x greater than 2, the function g(x) increases at a constant rate of 1 unit per x unit.
Now, let's go through each question:
A. Is it possible, impossible, or certain that g is continuous at x=2? Explain.
To determine if g is continuous at x = 2, we need to see if the graph of g'(x) has a "jump" or "hole" at x = 2. Looking at the graph, we see that the two segments meet at the point (2, -1) with a different slope on each side. This indicates that g(x) has a jump discontinuity at x = 2. Therefore, it is impossible for g to be continuous at x = 2.
B. Is it possible, impossible, or certain that g(2)=0? Explain.
Since g(x) has a jump discontinuity at x = 2, there is no defined value for g(2). Remember that the function g(x) is defined differently on each side of x = 2. On the left side, g(2) would be 3, and on the right side, g(2) would be -1. Therefore, it is impossible for g(2) to be 0.
C. Is it possible, impossible, or certain that g has a vertical asymptote at x = 2? Explain.
The concept of a vertical asymptote is related to a function approaching infinity or negative infinity as x approaches a certain value. Looking at the graph, we don't see any indication of the function g(x) approaching infinity or negative infinity as x approaches 2. Additionally, vertical asymptotes are often seen in functions involving rational expressions, which is not the case here. Therefore, it is impossible for g to have a vertical asymptote at x = 2.
D. Is it possible, impossible, or certain that g has a jump discontinuity at x = 2?
As mentioned before, looking at the graph of g'(x), we observe that there is a jump from one linear segment to the other at x = 2. This jump in slope indicates a jump discontinuity in g(x) at x = 2. Therefore, it is possible for g to have a jump discontinuity at x = 2.
E. Is it possible, impossible, or certain that g has a removable discontinuity at x = 2?
A removable discontinuity occurs when there is a hole in the graph that can be filled to make the function continuous at that point. In the case of g(x), there is a jump discontinuity at x = 2. Since a jump discontinuity cannot be "removed" by filling a hole, it is impossible for g to have a removable discontinuity at x = 2.
In summary, the correct answers are:
A) Impossible
B) Impossible
C) Impossible
D) Possible
E) Impossible