Let f be the function defined by
f(x) =
x^2 if x<0
square root x if 0 ≤ x <1
2-x if 1 ≤ x< 2
x-3 if x ≥ 2
For what numbers x is f not continuos?
1 only
2 only
0 and 2 only
1 and 2 only
0 is ok f(0) = 0
1 is ok, f(1) = 1
2 is NOT ok, f(2) = 0 and -1
If 0 and 1 is okay which answer would I Pick? It wouldnt be C or D
Nevermind it says not Continuous. So do I just plug in these numbers to each equation? Like Im confused how to solve it
If you get more than one value of f(x) for any x, the value changes instantly there. That is discontinuous. so if x = 2, tragedy.
To determine where the function f is not continuous, we need to identify the points where there could potentially be a discontinuity.
A function is continuous at a point if the limit of the function as x approaches that point exists and is equal to the value of the function at that point.
Let's examine each piece of the function individually and determine if there are any potential points of discontinuity:
1. For x < 0, f(x) = x^2. This is a polynomial function, and polynomials are continuous for all values of x.
2. For 0 ≤ x < 1, f(x) = √x. This is a square root function. Square root functions are continuous for positive values of x, so there's no discontinuity in this interval.
3. For 1 ≤ x < 2, f(x) = 2 - x. This is a linear function, and linear functions are continuous for all values of x.
4. For x ≥ 2, f(x) = x - 3. Again, this is a linear function, so there's no discontinuity in this interval.
Based on the analysis of each piece of the function, there are no points of discontinuity within each interval. Therefore, f(x) is continuous for all values of x.
So, none of the options you listed (1 only, 2 only, 0 and 2 only, 1 and 2 only) are correct.