Question

Let f(x),show below be defined by f(x)={-2x-2 x belongs to (-infinity,-1),g(x) x belongs to A,3 x belongs to [2,infinity)}

Which of the following definition of g(x) and A would make f(x) continuous on R?
g(x)= (x^2)-1, A=(-1,2]
g(x)=(x^2)-1,A=(-1,2)
g(x)=(x^2)+1,A=(-1,2)
g(x)=(x^2)-1,A=[-1,2)

Answer 1

Explanation:

For the function f(x) to be continuous on the entire real number line R, we need to ensure that it is continuous at the points where it transitions from one piecewise-defined function to another. In this case, we have transitions at x = -1 and x = 2. Therefore, we need to check the continuity conditions at these points.

Here are the criteria for continuity at these points:

1. The limit of f(x) as x approaches -1 from the left (-1-) must be equal to the limit of f(x) as x approaches -1 from the right (-1+).

2. The limit of f(x) as x approaches 2 from the left (2-) must be equal to the limit of f(x) as x approaches 2 from the right (2+).

3. The value of f(x) at x = -1 must be equal to the value of f(x) at x = -1.

4. The value of f(x) at x = 2 must be equal to the value of f(x) at x = 2.

Now, let's evaluate these criteria for each given choice of g(x) and A:

Choice 1:

g(x) = (x^2) - 1

A = (-1, 2]

- Limit at x = -1 from the left: (-2) - 2 = -4

- Limit at x = -1 from the right: (0) - 1 = -1

- Limit at x = 2 from the left: 3 - 2 = 1

- Limit at x = 2 from the right: 3 - 2 = 1

- Value at x = -1: (-1)^2 - 1 = 0

- Value at x = 2: (2^2) - 1 = 3

Choice 2:

g(x) = (x^2) - 1

A = (-1, 2)

- Limit at x = -1 from the left: (-2) - 2 = -4

- Limit at x = -1 from the right: (0) - 1 = -1

- Limit at x = 2 from the left: 3 - 2 = 1

- Limit at x = 2 from the right: 3 - 2 = 1

- Value at x = -1: (-1)^2 - 1 = 0

- Value at x = 2: (2^2) - 1 = 3

Choice 3:

g(x) = (x^2) + 1

A = (-1, 2)

- Limit at x = -1 from the left: (-2) + 1 = -1

- Limit at x = -1 from the right: (0) + 1 = 1

- Limit at x = 2 from the left: 3 + 1 = 4

- Limit at x = 2 from the right: 3 + 1 = 4

- Value at x = -1: (-1)^2 + 1 = 2

- Value at x = 2: (2^2) + 1 = 5

Choice 4:

g(x) = (x^2) - 1

A = [-1, 2)

- Limit at x = -1 from the left: (-2) - 2 = -4

- Limit at x = -1 from the right: (0) - 1 = -1

- Limit at x = 2 from the left: 3 - 2 = 1

- Limit at x = 2 from the right: 3 - 2 = 1

- Value at x = -1: (-1)^2 - 1 = 0

- Value at x = 2: (2^2) - 1 = 3

Based on the continuity criteria, choice 2, where g(x) = (x^2) - 1 and A = (-1, 2), makes f(x) continuous on R because it satisfies all four conditions.