Describe the interval(s)on which the function f(x)=(x-11/x^2-121) is continuous.
f(x) = 1/(x+11) where x is not 11 or -11. It looks like it ought to be true for x=11 as well, except that at x=11, we have f(x)=0/0.
So, the function is continuous on
(-oo,-11) (-11,11) and (11,+oo)
While the limit from the left and right at x=11 is 1, the function is not continuous at x=11 because f(11) is not defined.
To determine the intervals on which a given function is continuous, we need to consider two main factors: the domain of the function and any points of discontinuity.
First, let's identify the domain of the function f(x) = (x - 11) / (x^2 - 121). The denominator cannot be zero since division by zero is undefined. So, we need to find the values of x that make the denominator equal to zero:
x^2 - 121 = 0
This equation can be factored as (x - 11)(x + 11) = 0. Setting each factor equal to zero, we get x - 11 = 0 or x + 11 = 0. Solving these equations, we find x = 11 or x = -11.
Thus, the domain of the function is all real numbers except x = 11 and x = -11. We can write the domain as: (-∞, -11) ∪ (-11, 11) ∪ (11, ∞).
To find the intervals on which the function is continuous, we need to analyze any potential points of discontinuity at x = 11 and x = -11. Since these are points where the function is undefined, we need to check if the limit of the function exists as x approaches these points from both the left and right sides.
For x < -11, the function is well defined, and the limit exists as x approaches -11 from the left side. Similarly, for x > 11, the function is well defined, and the limit exists as x approaches 11 from the right side.
Therefore, the function is continuous for all x < -11 and x > 11.
In summary, the function f(x) = (x - 11) / (x^2 - 121) is continuous on the intervals (-∞, -11) and (11, ∞).