Continuity: Learn It 4

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Continuity: Learn It 4

Determining the Input Values for Which a Function Is Discontinuous

Now that we can identify continuous functions, jump discontinuities, and removable discontinuities, we will look at more complex functions to find discontinuities. Here, we will analyze a piecewise function to determine if any real numbers exist where the function is not continuous. A piecewise function may have discontinuities at the boundary points of the function as well as within the functions that make it up.

To determine the real numbers for which a piecewise function composed of polynomial functions is not continuous, recall that polynomial functions themselves are continuous on the set of real numbers. Any discontinuity would be at the boundary points. So we need to explore the three conditions of continuity at the boundary points of the piecewise function.

How To: Given a piecewise function, determine whether it is continuous at the boundary points.

For each boundary point [latex]a[/latex] of the piecewise function, determine the left- and right-hand limits as [latex]x[/latex] approaches [latex]a[/latex], as well as the function value at [latex]a[/latex].
Check each condition for each value to determine if all three conditions are satisfied.
Determine whether each value satisfies condition 1: [latex]f\left(a\right)[/latex] exists.
Determine whether each value satisfies condition 2: [latex]\underset{x\to a}{\mathrm{lim}}f\left(x\right)[/latex] exists.
Determine whether each value satisfies condition 3: [latex]\underset{x\to a}{\mathrm{lim}}f\left(x\right)=f\left(a\right)[/latex].
If all three conditions are satisfied, the function is continuous at [latex]x=a[/latex]. If any one of the conditions fails, the function is not continuous at [latex]x=a[/latex].

Determine whether the function [latex]f[/latex] is discontinuous for any real numbers.

[latex]fx=\begin{cases}x+1, \hfill& x<2 \\ 3, \hfill& 2\leq x<4 \\ x^{2}-11, \hfill& x\geq 4\end{cases}[/latex]

The piecewise function is defined by three functions, which are all polynomial functions, [latex]f\left(x\right)=x+1[/latex] on [latex]x<2[/latex], [latex]f\left(x\right)=3[/latex] on [latex]2\le x<4[/latex], and [latex]f\left(x\right)={x}^{2}-5[/latex] on [latex]x\ge 4[/latex]. Polynomial functions are continuous everywhere. Any discontinuities would be at the boundary points, [latex]x=2[/latex] and [latex]x=4[/latex]. At [latex]x=2[/latex], let us check the three conditions of continuity. Condition 1:

[latex]\begin{align}&f\left(2\right)=3\\ &\Rightarrow \text{Condition 1 is satisfied}.\end{align}[/latex]

Condition 2: Because a different function defines the output left and right of [latex]x=2[/latex], does [latex]\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)?[/latex]

Left-hand limit: [latex]\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{-}}{\mathrm{lim}}\left(x+1\right)=2+1=3[/latex]
Right-hand limit: [latex]\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}3=3[/latex]

Because [latex]3=3[/latex] , [latex]\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)[/latex]

[latex]\Rightarrow \text{Condition 2 is satisfied}[/latex].

Condition 3:

[latex]\begin{align}&\underset{x\to 2}{\mathrm{lim}}f\left(x\right)=3=f\left(2\right)\\ &\Rightarrow \text{Condition 3 is satisfied}.\end{align}[/latex]

Because all three conditions are satisfied at [latex]x=2[/latex], the function [latex]f\left(x\right)[/latex] is continuous at [latex]x=2[/latex].

At [latex]x=4[/latex], let us check the three conditions of continuity.

Condition 2: Because a different function defines the output left and right of [latex]x=4[/latex], does [latex]\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)?[/latex]

Left-hand limit: [latex]\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {4}^{-}}{\mathrm{lim}}3=3[/latex]
Right-hand limit: [latex]\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {4}^{+}}{\mathrm{lim}}\left({x}^{2}-11\right)={4}^{2}-11=5[/latex]

Because [latex]3\ne 5[/latex] , [latex]\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)[/latex], so [latex]\underset{x\to 4}{\mathrm{lim}}f\left(x\right)[/latex] does not exist.

[latex]\Rightarrow \text{Condition 2 fails}[/latex].

Because one of the three conditions does not hold at [latex]x=4[/latex], the function [latex]f\left(x\right)[/latex] is discontinuous at [latex]x=4[/latex].

Analysis of the Solution

At [latex]x=4[/latex], there exists a jump discontinuity. Notice that the function is continuous at [latex]x=2[/latex].

Graph of a piecewise function that has disconuity at (4, 3). — Graph is continuous at [latex]x=2[/latex] but shows a jump discontinuity at [latex]x=4[/latex].

Determine where the function

[latex]f\left(x\right)=\begin{cases}\frac{\pi x}{4}, &x<2 \\ \frac{\pi}{x}, &2\leq x \leq 6 \\ 2\pi x, &x>6\end{cases}[/latex]

is discontinuous.

Determining Whether a Function Is Continuous

To determine whether a piecewise function is continuous or discontinuous, in addition to checking the boundary points, we must also check whether each of the functions that make up the piecewise function is continuous.

How To: Given a piecewise function, determine whether it is continuous.

Determine whether each component function of the piecewise function is continuous. If there are discontinuities, do they occur within the domain where that component function is applied?
For each boundary point [latex]x=a[/latex] of the piecewise function, determine if each of the three conditions hold.

Determine whether the function below is continuous. If it is not, state the location and type of each discontinuity.

[latex]fx=\begin{cases}sin\left(x\right), \hfill& x<0 \\ x^{3}, \hfill& x>0\end{cases}[/latex]