Question 11.3.2: Determining Where a Piecewise Function Is Discontinuous Dete...

First, let’s determine whether each of the three pieces of f is continuous. The first piece, f(x) = x + 2, is a linear function; it is continuous at every number x. The second piece, f(x) = 2, a constant function, is continuous at every number x. And the third piece, f(x) = x² + 2, a polynomial function, is also continuous at every number x. Thus, these three functions, a linear function, a constant function, and a polynomial function, can be graphed without lifting a pencil from the paper. However, the pieces change at x = 0 and at x = 1. Is it necessary to lift a pencil from the paper when graphing f at these values? It appears that we must investigate continuity at 0 and at 1.

To determine whether the function is continuous at 0, we check the conditions for continuity with a = 0.

Condition 1 f is defined at a. Is f(0) defined? Because a = 0, we use the first line of the piecewise function, where x ≤ 0.

f(x) = x + 2       This is the function’s equation for x ≤ 0, which includes x = 0.

f(0) = 0 + 2        Replace x with 0.

= 2

Because f(0) is a real number, 2, f(0) is defined.

Condition 2 \underset{x\rightarrow a}{\lim }f(x) exists. Does \underset{x\rightarrow 0}{\lim }f(x) exist? To answer this question, we look at the values of f(x) when x is close to 0. Let us investigate the left- and right-hand limits. If these limits are equal, then \underset{x\rightarrow 0}{\lim }f(x) exists. To find \underset{x\rightarrow 0}{\lim }f(x), the left-hand limit, we look at the values of f(x) when x is close to 0 but less than 0. Because x is less than 0, we use the first line of the piecewise function, f(x) = x + 2 if x ≤ 0. Thus,

\underset{x\rightarrow 0^-}{\lim }f(x)=\underset{x\rightarrow 0^-}{\lim }(x + 2) = 0 + 2 = 2.

To find \underset{x\rightarrow 0^+}{\lim }f(x), the right-hand limit, we look at the values of f(x) when x is close to 0 but greater than 0. Because x is greater than 0, we use the second line of the piecewise function, f(x) = 2 if 0 < x ≤ 1. Thus,

\underset{x\rightarrow 0^+}{\lim }f(x)=\underset{x\rightarrow 0^+}{\lim } 2 = 2.

Because the left- and right-hand limits are both equal to 2, \underset{x\rightarrow 0}{\lim }f(x)= 2. Thus, we
see that \underset{x\rightarrow 0}{\lim }f(x) exists.

Condition 3 \underset{x\rightarrow a}{\lim }f(x)= f (a)  Does \underset{x\rightarrow 0}{\lim }f(x)= f (0)? We found that \underset{x\rightarrow 0}{\lim }f(x)= 2 and f(0) = 2. This means that as x gets closer to 0, the corresponding values of f(x) get closer to the function value at 0: \underset{x\rightarrow 0}{\lim }f(x)= f (0).

Because the three conditions are satisfied, we conclude that f is continuous at 0.

Now we must determine whether the function is continuous at 1, the other value of x where the pieces change. We check the conditions for continuity with a = 1.

Condition 1 f is defined at a. Is f(1) defined? Because a = 1, we use the second line of the piecewise function, where 0 < x ≤ 1.

f(x) = 2                     This is the function’s equation for 0 < x ≤ 1, which                                                             includes x = 1.

f(1) = 2                     Replace x with 1.

Because f(1) is a real number, 2, f(1) is defined.

Condition 2 \underset{x\rightarrow a}{\lim }f(x) exists. Does \underset{x\rightarrow 1}{\lim }f(x) exist? We investigate left- and righthand limits as x approaches 1. To find \underset{x\rightarrow 1^-}{\lim }f(x), the left-hand limit, we look at values of f(x) when x is close to 1 but less than 1. Thus, we use the second line of the piecewise function, f(x) = 2 if 0 < x ≤ 1. The left-hand limit is

\underset{x\rightarrow 1^-}{\lim }f(x) =\underset{x\rightarrow 1^-}{\lim }2=2.

To find \underset{x\rightarrow 1^+}{\lim }f(x), the right-hand limit, we look at values of f(x) when x is close to 1 but greater than 1. Thus, we use the third line of the piecewise function, f(x) = x² + 2 if x > 1. The right-hand limit is

\underset{x\rightarrow 1^+}{\lim }f(x)=\underset{x\rightarrow 1^+}{\lim } (x² + 2) = 1² + 2 = 3.

The left- and right-hand limits are not equal: \underset{x\rightarrow 1^-}{\lim }f(x)=2 and \underset{x\rightarrow 1^+}{\lim }f(x)=3.

This means that \underset{x\rightarrow 1}{\lim }f(x) does not exist.

Because one of the three conditions is not satisfied, we conclude that f is not continuous at 1.

In summary, the given function is discontinuous at 1 only. The graph of f, shown in Figure 11.14, illustrates this conclusion.