Let f(x)=|x+1|/(x^(2)-1). Is f continuous? If not,classify any points of discontinuity as infinite, jump, or removable. Sketch the graph of y=f(x) for -4<=x<=4
since x^2-1=0 at x = ±1, f is not continuous.
For any other values,
f(x) = 1/(x-1) when x > -1
f(x) = -1/(x-1) for x < -1
evaluate the left and right limits near 1 and -1 to determine the nature of the discontinuities.
To determine if the function f(x) = |x+1|/(x^(2)-1) is continuous, we need to check for continuity at various points and intervals.
First, let's find the points of discontinuity by identifying any values of x where the function is undefined or where there are potential discontinuities.
1. Undefined points:
The function is undefined when the denominator of the fraction is equal to zero. In this case, x^2 - 1 = 0. Solving this equation, we find that x = -1 and x = 1 are the points where the function is undefined.
2. Jump discontinuity:
To check for a jump discontinuity, we need to evaluate the function at each side of a potential point of discontinuity. Let's consider x approaching -1 and 1 separately.
a) As x approaches -1 from the left side (x < -1), |x+1| is equal to -x-1, and the function becomes f(x) = (-x-1)/(x^2-1).
b) As x approaches -1 from the right side (x > -1), |x+1| is equal to x+1, resulting in f(x) = (x+1)/(x^2-1).
Comparing the limits from both sides, we find that they are not equal, which suggests a jump discontinuity at x = -1.
Following the same steps for x = 1, we get a similar result. Therefore, there is also a jump discontinuity at x = 1.
3. Removable discontinuity:
To identify a removable discontinuity, we need to verify if the function can be modified or simplified to fill the gap at a particular point. Let's examine x = -1 and x = 1.
At x = -1:
Using the fact that f(x) = (-x-1)/(x^2-1) when x < -1 and f(x) = (x+1)/(x^2-1) when x > -1, we can rewrite the function as f(x) = (-x-1)/(x+1)(x-1).
We can simplify further by canceling the common factor of x + 1 in both the numerator and denominator, resulting in f(x) = (-1)/(x-1).
At x = 1:
Using the same process, we find that f(x) simplifies to f(x) = 1/(x+1) when x < 1 and f(x) = (x+1)/(x+1)(x-1) when x > 1.
Cancelling the common factor of x + 1 in both the numerator and denominator, we obtain f(x) = 1/(x-1).
As we can modify the function at both x = -1 and x = 1 to fill the gap, these points represent removable discontinuities.
Now, let's sketch the graph of f(x) for -4 ≤ x ≤ 4, keeping in mind the points of discontinuity and type of discontinuity identified.
First, plot the points -1 and 1 on the x-axis and draw an open circle at these points to represent the jump discontinuities. Then, draw a dashed line to indicate a gap between the two parts of the graph at each of these points.
Next, plot the points -1 and 1 again, but this time with closed circles to represent the removable discontinuities. Connect the two branches of the graph using a solid line to fill the gap at each of these points.
Finally, plot the remaining points within the given range of -4 to 4 by calculating the function value for each x-value within that interval. Connect all the points with a smooth curve, keeping in mind the behavior of each portion of the graph based on the function equation in those ranges.
By following these steps, you can sketch the graph of y = f(x) for -4 ≤ x ≤ 4, considering the identified points of discontinuity and their types.