suppose f(x) = {x-1/x^4-1 if x <= 2 and |x-3|/x-3 if 2<x
Identify any points of discontinuity, and determine (giving reasons) if they are removable, infinite (essential), or jump discontinuities.
(x-1)/(x^4-1) = 1/((x+1)(x^2+1)) for x≠1
There is a hole at x=1
|x-3|/(x-3) = 1 if x>3
= -1 if x<3
So there is a hole at x=3
lim(x→2-) f(x) = 1/15
lim(x→2+) f(x) = -1
So there is a jump discontinuity at x=2
To identify the points of discontinuity of the given function f(x) = {x-1/x^4-1 if x <= 2 and |x-3|/x-3 if 2 < x, we need to examine the behavior of the function at different points in its domain.
1. At x = 2:
For x <= 2, the function is defined as (x-1)/(x^4-1). At x = 2, the denominator becomes 2^4 - 1 = 15, which is not equal to zero. Therefore, there is no discontinuity at x = 2.
2. At x = 3:
For x > 2, the function is defined as |x-3|/(x-3). At x = 3, the denominator becomes 3 - 3 = 0. This creates a potential point of discontinuity.
To determine the type of discontinuity at x = 3, we need to evaluate the limit of the function as x approaches 3 from the left and right sides.
From the left side (x < 3):
lim(x->3-) f(x) = lim(x->3-) (|x-3|)/(x-3)
= lim(x->3-) (x-3)/(x-3) (since |x-3| = x-3 when x < 3)
= lim(x->3-) 1
= 1
From the right side (x > 3):
lim(x->3+) f(x) = lim(x->3+) (|x-3|)/(x-3)
= lim(x->3+) (-(x-3))/(x-3) (since |x-3| = -(x-3) when x > 3)
= lim(x->3+) -1
= -1
Since the limit from the left (1) is not equal to the limit from the right (-1), this is a non-removable (essential) discontinuity at x = 3.
In summary:
- The function f(x) has no discontinuity at x = 2.
- There is an essential discontinuity at x = 3.
To identify points of discontinuity in the given function f(x), we need to examine the behavior of the function at specific values of x and check if there are any breaks or jumps in the graph.
Let's consider the different cases separately:
1. For x ≤ 2: In this case, the function is defined as f(x) = (x-1)/(x^4-1).
To identify any points of discontinuity, we need to determine if there are any values of x where the denominator becomes zero, leading to an undefined value. In this case, the denominator (x^4-1) will be equal to zero when x = 1 or x = -1. Hence, we have potential points of discontinuity at x = 1 and x = -1.
2. For 2 < x: In this case, the function is defined as f(x) = |x-3|/(x-3).
Since the function involves an absolute value, it's important to check if there are any values of x that make the denominator (x-3) equal to zero. In this case, x = 3 would lead to an undefined value. Thus, we have a potential point of discontinuity at x = 3.
Now, let's examine each potential point of discontinuity and determine if they are removable, infinite (essential), or jump discontinuities.
1. x = 1 and x = -1 (for x ≤ 2): At these points, the function is well-defined without any issues. Hence, they are not points of discontinuity.
2. x = 3 (for 2 < x): Here, we have an essential or infinite discontinuity. When x approaches 3 from the left (x < 3), the function value is negative infinity, and when x approaches 3 from the right (x > 3), the function value is positive infinity. Therefore, we have an essential or infinite discontinuity at x = 3.
In summary, the function f(x) has one point of essential or infinite discontinuity at x = 3.