Describe the intervals on which the following function is continuous. If not continuous, state "where" the discontinuity occurs and state the TYPE of discontinuity.
f(x)= (x^2-x-12)/(x^2-6x+8)
I just don't understand how to find the intervals. Thanks~
f(x)= (x^2-x-12)/(x^2-6x+8)
= (x-4)(x+3)/( (x-2)(x-4)
= (x+3)/(x-2) , x not= 4
So the function is discontinuous at x=4 and at x = 2
with a "hole" at (4, 7/2)
and a vertical asymptote at x = 2
So continuous for
x<2 OR x > 2, x not= 4
To determine the intervals on which a function is continuous, we need to consider the domain of the function and identify any potential points of discontinuity.
The given function f(x) = (x^2 - x - 12)/(x^2 - 6x + 8) can be simplified first. The numerator can be factored as (x - 4)(x + 3), and the denominator as (x - 4)(x - 2). Thus, the function can be written as:
f(x) = (x - 4)(x + 3)/(x - 4)(x - 2)
Now, we can see that (x - 4) appears both in the numerator and denominator. This suggests that there might be a discontinuity at x = 4.
To investigate further, let's calculate the limit of f(x) as x approaches 4 from either side:
lim(x->4-) f(x) = lim(x->4-) [(x - 4)(x + 3)]/[(x - 4)(x - 2)]
= lim(x->4-) (x + 3)/(x - 2)
= 7/2
lim(x->4+) f(x) = lim(x->4+) [(x - 4)(x + 3)]/[(x - 4)(x - 2)]
= lim(x->4+) (x + 3)/(x - 2)
= 7/2
Since both the right-hand and left-hand limits are equal, the limit of f(x) as x approaches 4 exists. This means that the potential point of discontinuity at x = 4 is removable. So, we can remove the common factor (x - 4) from both the numerator and denominator.
After removing the common factor, the simplified expression for f(x) is:
f(x) = (x + 3)/(x - 2)
Now, for this simplified function, we can see that there are no other points of discontinuity in the domain. Therefore, the function f(x) = (x^2 - x - 12)/(x^2 - 6x + 8) is continuous on all real numbers except x = 4.
To determine the intervals on which a function is continuous, you need to consider two factors:
1. Denominator not equal to zero: First, find the values of x that make the denominator equal to zero, as these values will create potential points of discontinuity. In this case, the denominator is (x^2 - 6x + 8), so let's find when it equals zero:
(x^2 - 6x + 8) = 0
Factoring: (x - 2)(x - 4) = 0
So, x = 2, x = 4 are the values that make the denominator zero.
2. Simplify the expression: Simplify the expression, excluding the values of x found in the denominator. This will help identify any potential points of discontinuity caused by a factor that should be canceled out, but isn't.
Now let's analyze the given function f(x) = (x^2 - x - 12) / (x^2 - 6x + 8) step by step:
1. Numerator: The numerator is a quadratic equation, and it doesn't have any specific values that make it equal to zero.
2. Denominator: We've already found the values of x that make the denominator equal to zero, which are x = 2 and x = 4.
3. Simplify the expression:
f(x) = (x^2 - x - 12) / (x^2 - 6x + 8)
The denominator can be factored as (x - 2)(x - 4), but neither of the factors can be simplified further.
Now, we have all the necessary information to determine the intervals of continuity:
1. Interval (-∞, 2): In this interval, exclude x = 2 as it makes the denominator zero. Therefore, the function is continuous in this interval.
2. Interval (2, 4): In this interval, the denominator is positive, and the numerator remains finite. Hence, the function is continuous in this interval as well.
3. Interval (4, +∞): In this interval, exclude x = 4 as it makes the denominator zero. Similar to the first interval, the function is continuous here.
To summarize, the function f(x) = (x^2 - x - 12) / (x^2 - 6x + 8) is continuous for all values of x except x = 2 and x = 4, as these values cause the denominator to be zero. These points where the function is discontinuous are called "removable" discontinuities, as they can be canceled out by simplification.