Find the discontinuities of f(x)= ((x^2)+5x+6)/((x^2)-4)and categorize them as removable or non removable.
Start by finding the zeroes of the denominator.
since x^2+5x+6 = (x+2)(x+3)
and x^2-4 = (x+2)(x-2)
f(x) = (x+3)/(x-2) for x ≠ -2
so, x = -2 is a removable discontinuity,
x = 2 is a nonremovable discontinuity
extra credit: how do we remove the discontinuity at x = -2?
To find the discontinuities of the function f(x) = ((x^2) + 5x + 6)/((x^2) - 4), we need to identify the values of x that would make the denominator zero (since division by zero is undefined).
First, let's find the values of x that make the denominator zero by setting (x^2) - 4 = 0 and solving for x:
(x^2) - 4 = 0
(x - 2)(x + 2) = 0
The solutions to this equation are x = 2 and x = -2. Therefore, the function f(x) has two potential discontinuities at x = 2 and x = -2.
To categorize these discontinuities as removable or non-removable, we need to examine the behavior of the function at these points.
1. For x = 2:
To determine if this is a removable discontinuity, we need to evaluate the limit of the function as x approaches 2.
lim(x→2) ((x^2) + 5x + 6)/((x^2) - 4)
By substituting x = 2 into the function, we get:
f(2) = ((2^2) + 5(2) + 6)/((2^2) - 4)
f(2) = (4 + 10 + 6)/(4 - 4)
f(2) = 20/0
Since the denominator is zero, the function is undefined at x = 2. This means that the discontinuity at x = 2 is non-removable.
2. For x = -2:
Similarly, we need to evaluate the limit of the function as x approaches -2.
lim(x→-2) ((x^2) + 5x + 6)/((x^2) - 4)
By substituting x = -2 into the function, we get:
f(-2) = ((-2^2) + 5(-2) + 6)/((-2^2) - 4)
f(-2) = (4 - 10 + 6)/(4 - 4)
f(-2) = 0/0
Again, the denominator is zero, resulting in an undefined value. Therefore, the discontinuity at x = -2 is also non-removable.
To summarize, the function f(x) = ((x^2) + 5x + 6)/((x^2) - 4) has two discontinuities at x = 2 and x = -2, both of which are non-removable.
To find the discontinuities of a function, we should look for values of x where the function is undefined, or where the denominator becomes zero. In this case, the denominator is (x^2 - 4). So, we need to find the values of x that would make the denominator equal to zero.
To solve the equation x^2 - 4 = 0, we can factor the quadratic expression. Factoring is possible because it is a difference of squares.
(x^2 - 4) = (x - 2)(x + 2)
Setting the expression equal to zero, we have:
(x - 2)(x + 2) = 0
This equation has two solutions: x = 2 and x = -2. These are the values of x that make the denominator of the function equal to zero.
Now, we need to categorize these discontinuities as removable or non-removable. A removable discontinuity occurs if the function can be redefined at that point to make it continuous. A non-removable discontinuity, also known as an essential or infinite discontinuity, cannot be fixed by redefining the function.
To determine if these discontinuities are removable or non-removable, we need to check the behavior of the function as x approaches these values.
For x = 2:
As x approaches 2, the numerator of the function becomes (2^2 + 5*2 + 6) = 4 + 10 + 6 = 20, and the denominator becomes zero. This means the function approaches infinity or negative infinity. Hence, the discontinuity at x = 2 is non-removable.
For x = -2:
Similarly, as x approaches -2, the numerator of the function becomes (-2^2 + 5*(-2) + 6) = 4 - 10 + 6 = 0, and the denominator becomes zero. This means the function approaches zero. Hence, the discontinuity at x = -2 is removable because we can redefine the function at that point to make it continuous.
To summarize:
- The value x = 2 corresponds to a non-removable discontinuity.
- The value x = -2 corresponds to a removable discontinuity.