g(x)= {0 if x<0
{x^2-2x if 0≤x≤2
{2 if x>2
Find all values x=a where the function is discontinuous.
Is the answer 0 and 2 or just 2?
x<0 , 0 ≤ x ≤ 2 , x > 2
all the x's "link" properly with no gaps, everybody is accounted for.
So there are no values of x where the function is discontinuous
( that is, for every x that I choose, I can find a y value)
g(x)= {0 if x<0
{x^3-3x if 0≤x≤3
{3 if x>3
If it was like this one. Then what? Same
so answer would be none or nowhere.
To determine where the function g(x) is discontinuous, we need to identify any values of x where the function encounters a jump or a hole in its graph.
In this case, we can see that g(x) has three different definitions for different intervals of x.
First, let's check for discontinuity at x = 0. The definition of g(x) for x less than 0 is 0, and for x greater than 0 and less than or equal to 2 is x^2 - 2x. Therefore, to check for a jump or hole, we need to calculate the left-hand limit and the right-hand limit at x = 0.
Left-hand limit of g(x) as x approaches 0 (from the negative side):
lim_(x→0-) g(x) = 0
Right-hand limit of g(x) as x approaches 0 (from the positive side):
lim_(x→0+) g(x) = (0)^2 - 2(0) = 0
Since the left-hand limit and right-hand limit are equal, g(x) is continuous at x = 0.
Next, let's check for discontinuity at x = 2. The definition of g(x) for x greater than 2 is 2. Therefore, we need to calculate the left-hand limit and the right-hand limit at x = 2.
Left-hand limit of g(x) as x approaches 2 (from the negative side):
lim_(x→2-) g(x) = (2)^2 - 2(2) = 0
Right-hand limit of g(x) as x approaches 2 (from the positive side):
lim_(x→2+) g(x) = 2
Since the left-hand limit and right-hand limit are not equal, g(x) is discontinuous at x = 2.
Therefore, the function g(x) is discontinuous only at x = 2.