Given
f(x) = (x^4 + 17) / (6x^2 + x - 1)
Identify any points of discontinuity, and determine (giving reasons) if they are removable, infinite (essential), or jump discontinuities.
From the work that I have done so far, I know that there are two discontinuiities, one at x = -1/2 and another at x = 1/3. On my graphing calculator, it appears that they form an infinite (or essential) discontinuity. Is this correct?
That is correct. In order to have a removable discontinuity, you need f(x) = 0/0 which is undefined, but may have a limit. In this case, the numerator is never zero.
To determine the nature of the discontinuities at x = -1/2 and x = 1/3 for the function f(x) = (x^4 + 17) / (6x^2 + x - 1), we need to analyze the behavior of the function as x approaches these points.
1. Discontinuity at x = -1/2:
To analyze the behavior of f(x) as x approaches -1/2, substitute -1/2 into the function:
f(-1/2) = ((-1/2)^4 + 17) / (6(-1/2)^2 - 1/2 - 1)
= (1/16 + 17) / (6/4 - 1/2 - 1)
= (17/16 + 17) / (3/2 - 1/2 - 1)
= (17/16 + 17) / (3/2 - 3/2)
= (17/16 + 17) / 0
We can see that the denominator becomes zero, which indicates a potential point of discontinuity. To determine the nature of this discontinuity, we can analyze the limits of f(x) as x approaches -1/2 from both the left and right sides.
Left-hand limit:
lim(x->-1/2-) f(x) = lim(x->-1/2-) ((x^4 + 17) / (6x^2 + x - 1))
= lim(x->-1/2-) ((-x)^4 + 17) / (6(-x)^2 + (-x) - 1)
= lim(x->-1/2-) (x^4 + 17) / (6x^2 - x - 1)
= ( (-1/2)^4 + 17 ) / ( 6(-1/2)^2 - (-1/2) - 1 )
= (1/16 + 17) / (6/4 + 1/2 - 1)
= (17/16 + 17) / (3/2 - 1/2 - 1)
= (17/16 + 17) / (3/2 - 3/2)
= ( 17/16 + 17 ) / 0
Right-hand limit:
lim(x->-1/2+) f(x) = lim(x->-1/2+) ((x^4 + 17) / (6x^2 + x - 1))
= lim(x->-1/2+) ((-x)^4 + 17) / (6(-x)^2 + (-x) - 1)
= lim(x->-1/2+) (x^4 + 17) / (6x^2 - x - 1)
= ( (-1/2)^4 + 17 ) / ( 6(-1/2)^2 - (-1/2) - 1 )
= (1/16 + 17) / (6/4 + 1/2 - 1)
= (17/16 + 17) / (3/2 - 1/2 - 1)
= (17/16 + 17) / (3/2 - 3/2)
= ( 17/16 + 17 ) / 0
From the left-hand and right-hand limits, we see that in both cases, the numerator does not approach a finite limit as x approaches -1/2, whereas the denominator approaches zero. Therefore, we can conclude that the discontinuity at x = -1/2 is an infinite (essential) discontinuity.
2. Discontinuity at x = 1/3:
To analyze the behavior of f(x) as x approaches 1/3, substitute 1/3 into the function:
f(1/3) = ((1/3)^4 + 17) / (6(1/3)^2 + 1/3 - 1)
= (1/81 + 17) / (6/9 + 1/3 - 1)
= (17/81 + 17) / (2/3 + 1/3 - 1)
= (17/81 + 17) / (2/3)
We can see that the denominator does not become zero at x = 1/3. Therefore, there is no discontinuity at x = 1/3.
In conclusion, there is an infinite (essential) discontinuity at x = -1/2, but there is no discontinuity at x = 1/3.
To identify points of discontinuity in a function, we need to find the values of x where the function is not defined or where it behaves differently than the rest of the graph.
In the given function f(x) = (x^4 + 17) / (6x^2 + x - 1), the denominator (6x^2 + x - 1) cannot be equal to zero, as division by zero is undefined. So, we will solve the equation 6x^2 + x - 1 = 0 to find the values of x that lead to discontinuities.
Using the quadratic formula, we have x = (-b ± √(b^2 - 4ac)) / (2a), where a = 6, b = 1, and c = -1:
x = (-1 ± √(1^2 - 4 * 6 * -1)) / (2 * 6)
x = (-1 ± √(1 + 24)) / 12
x = (-1 ± √25) / 12
x = (-1 ± 5) / 12
This gives us two possible values for x: x = 4/12 = 1/3 and x = -6/12 = -1/2. These are the points where the function may have discontinuities.
Now we need to determine whether these discontinuities are removable, infinite (essential), or jump discontinuities.
To determine if a discontinuity is removable, we evaluate the function at the point of discontinuity and check if it can be defined or "filled in" to make it continuous. If the limit of the function as it approaches the point exists and is finite, it is a removable discontinuity.
For the point x = 1/3:
f(1/3) = [(1/3)^4 + 17] / [6(1/3)^2 + 1/3 - 1]
f(1/3) = [1/81 + 17] / [6/9 + 3/9 - 9/9]
f(1/3) = [1/81 + 17] / [6/9 - 5/9]
f(1/3) = [1/81 + 17] / [1/9]
f(1/3) = (1 + 17/81) * 9/1
f(1/3) = (81 + 17) / 81
f(1/3) = 98 / 81
Since the function can be defined at x = 1/3, the discontinuity at this point is not removable.
For the point x = -1/2:
f(-1/2) = [(-1/2)^4 + 17] / [6(-1/2)^2 - 1/2 - 1]
f(-1/2) = [1/16 + 17] / [6/4 - 2/4 - 4/4]
f(-1/2) = [1/16 + 17] / [6/4 - 6/4]
f(-1/2) = [1/16 + 17] / [0/4]
f(-1/2) = ∞ / 0
The limit of the function as it approaches x = -1/2 is infinity, and the denominator becomes zero. This indicates an infinite (essential) discontinuity. Therefore, the point x = -1/2 is an infinite discontinuity.
In conclusion, based on the calculations, the discontinuity at x = -1/2 is an infinite (essential) discontinuity, while the discontinuity at x = 1/3 is not removable.