Matthew P. answered • 09/10/19
Computer Science Student Who Loves Teaching Math
Hi there!
A function is considered continuous at some x if (1) the limit at that x is the same from both sides, and (2) the limit is equal to the function's value for that x. All the expressions in this piecewise function are well-defined for all inputs of x, so we just need to check the limits at the values at which they change from one expression to the next: in particular, we need to consider x = 1 and x = 3.
Right off the bat, we can see that neither x = 2 nor x = 4 would have any reason to not be continuous, so we can rule out those answers right away.
For the two values we are worried about:
x = 1: if we look at the three expressions at and around x = 1,
x2 + 2x, x < 1
3, x = 1
x3 + x2 + x, x ∈ (1, 3)
we can see that all these expressions have well-defined values when x = 1. Therefore, a valid way to determine the limit at x = 1 is to evaluate each expression plugging in x = 1 for each:
limx→1- f(x) = 12 + 2(1) = 1 + 2 = 3
f(1) = 3 (given)
limx→1+ f(x) = 13 + 12 + 1 = 1 + 1 + 1 = 3
Since the limit from both sides is equal to the value of f(1), we can say that f is continuous at x = 1.
At this point, we could simply use process of elimination to claim that f is noncontinuous at x = 3, but in the spirit of learning let's see why f is not continuous at x = 3.
x = 3: if we look at the expressions at and around x = 3,
x3 + x2 + x, x ∈ (1, 3)
0, x = 3
2x + 1, x > 3
we see that, just like x = 1, these expressions are all well-defined at x = 3, so we can check for limits by just plugging in x = 3:
limx→3- f(x) = 33 + 32 + 3 = 27 + 9 + 3 = 39
f(3) = 0 (given)
limx→3+ f(x) = 2(3) + 1 = 7
We can see that all three of these values differ wildly from one another, so we say that f is noncontinuous at x = 3.
I hope this helped, and that you have a better understanding of checking for continuity. Let me know if you have any questions!
