Find the points at which the following function is discontinuous.Indicate the intervals at which the function is continuous.
f(x)= (x^2 if x<-1), (|x| if -1<x<=1), (3x+1 if 1<x)
looks like x = -1 is NOT included in any of the functions.
f is also discontinuous at x=1, since f(1) = 1, but f(1+h) = 4+h, which does not equal 1, no matter how small h is.
make that f(1+h) = 4 + 3h, but it's still nowhere near 1.
To find the points at which a function is discontinuous, we need to identify any values of x where the function has a jump, a hole, or an asymptote.
Let's analyze each part of the function separately to determine the points of discontinuity and the intervals where the function is continuous.
1. For x < -1:
Here, the function is defined as f(x) = x^2.
This region has no points of discontinuity since it's a smooth and continuous quadratic function.
2. For -1 < x ≤ 1:
This section of the function is defined as f(x) = |x|.
The absolute value function has a discontinuity at x = 0, where the graph changes direction or has a sharp turn. Therefore, f(x) = |x| is not continuous at x = 0.
3. For x > 1:
In this region, the function is defined as f(x) = 3x + 1.
The linear function f(x) = 3x + 1 is continuous for all values of x because it is a line without any jumps, holes, or asymptotes.
To summarize:
- The function f(x) = x^2 is continuous for all values of x.
- The function f(x) = |x| is discontinuous at x = 0.
- The function f(x) = 3x + 1 is continuous for all values of x.
Therefore, the points at which the given function f(x) is discontinuous are x = 0.
The intervals at which the function is continuous are x < -1, -1 < x ≤ 0, and x > 1.