Let g(x)={x^2-x-2 if x is not equal to 1
{0 if x=1
ok.
there is certainly a break at x=1.
Srry the question is incomplete...
please help with these questions
a. Is g(x) continuous at x=1
b. What type of discontinuity does it have
lim (x->1) = -2
g(1) = 0
So, there is a jump discontinuity. It is not removable, since g(1) is defined, but is not the same as the limit from both sides.
The function g(x) is defined as follows:
g(x) = x^2 - x - 2 if x is not equal to 1
g(x) = 0 if x = 1
To understand this function better, you can evaluate it for different values of x.
Let's start with x = 1:
- Since x is equal to 1, the function g(x) is defined as 0. So g(1) = 0.
Now let's evaluate g(x) for values of x other than 1. For example, let's try x = -2:
- Since x is not equal to 1, we use the first part of the definition: g(x) = x^2 - x - 2.
- Plugging in x = -2, we get g(-2) = (-2)^2 - (-2) - 2 = 4 + 2 - 2 = 4.
To summarize:
- g(1) = 0
- g(-2) = 4
You can evaluate g(x) for any other value of x by following the same process:
1. Check if x is equal to 1. If it is, the value of g(x) is 0.
2. If x is not equal to 1, use the expression x^2 - x - 2 to calculate g(x).
This way, you can find the value of g(x) for any given input.