Question

Let \(f(x)=\left\\{\begin{array}{ll}x^{2}-1 & x<3 \\ x+5 & x \geq 3\end{array}\right.\) Is \(f\) continuous everywhere?

Short answer

Yes, the function \(f\) is continuous everywhere.

Step-by-step solution

Define Continuity

A function is continuous at a point if the limit from the left equals the limit from the right, and both equal the function's value at that point. For our piecewise function, we need to check this condition at the boundary where the function changes: at \(x = 3\).

Evaluate Left-Hand Limit

For \(x < 3\), the function is \(f(x) = x^2 - 1\). Calculate the left-hand limit as \(x\) approaches 3: \[\lim_{{x \to 3^-}} f(x) = \lim_{{x \to 3}} (x^2 - 1) = 3^2 - 1 = 8.\]

Evaluate Right-Hand Limit

For \(x \geq 3\), the function is \(f(x) = x + 5\). Calculate the right-hand limit as \(x\) approaches 3: \[\lim_{{x \to 3^+}} f(x) = \lim_{{x \to 3}} (x + 5) = 3 + 5 = 8.\]

Evaluate Function Value at the Boundary

For \(x = 3\), we use the second case of the function, \(f(x) = x + 5\). Thus, \(f(3) = 3 + 5 = 8\).

Compare Limits and Function Value

Compare the left-hand limit, the right-hand limit, and the actual function value at \(x = 3\). All are equal: \(\lim_{{x \to 3^-}} f(x) = \lim_{{x \to 3^+}} f(x) = f(3) = 8\). Thus, \(f\) is continuous at \(x = 3\).

General Continuity Assessment

For points other than \(x = 3\), the function \(f\) is composed of polynomials \(x^2 - 1\) and \(x + 5\), both of which are continuous everywhere on their respective domains. Hence, \(f\) is continuous for all \(x\).

Key concepts

Piecewise Functions

A piecewise function is a type of function that is defined by different expressions or formulas over different parts of its domain. Think of it as a mathematical tool that suits different situations depending on the input value.
For example, the function given in our exercise is defined as:

• For values less than 3, it uses the formula: \(x^2 - 1\).
• For values equal to or greater than 3, it uses: \(x + 5\).

This setup allows the function to "switch" at the boundary point \(x = 3\), making piecewise functions very flexible.

Left-Hand Limit

When analyzing piecewise functions, it's essential to examine the behavior of the function as it approaches a particular point from the left. This is known as the left-hand limit.
For our function, we consider the limit as \(x\) approaches 3 from values less than 3. Using the formula \(x^2 - 1\), we calculate:

• \(\lim_{{x \to 3^-}} (x^2 - 1) = 8\).

This tells us what the function's output is getting close to as it moves closer to 3 from the left side.

Right-Hand Limit

Similarly, the right-hand limit examines how the function behaves as \(x\) approaches the point from the right. This is crucial in determining if the function is continuous.
For the function given, we use the formula \(x + 5\) for values equal to or greater than 3. Thus, we find:

• \(\lim_{{x \to 3^+}} (x+5) = 8\).

This result indicates what happens as \(x\) approaches 3 from the right-hand side.

Continuous Functions

A function is considered continuous at a point if certain conditions are met:

• The left-hand limit exists.
• The right-hand limit exists.
• Both limits equal the function's actual value at that point.

For the function in our example:

• The left-hand limit of 8 matches the right-hand limit of 8.
• The actual value \(f(3) = 8\).

Thus, the function is continuous at \(x = 3\), as well as at all other points since \(x^2 - 1\) and \(x + 5\) are continuous everywhere in their domains.

Limits in Calculus

Limits are a foundational concept in calculus that describe how a function behaves as it approaches a specific point. In the context of piecewise functions like ours, limits help determine continuity at boundary points.
Ratios, sums, or even piecewise functions may exhibit different behaviors as input values get infinitesimally close to a certain number. Calculating limits on both sides of a boundary point ensures a seamless transition between the pieces of a piecewise function.
Overall, limits are not just crucial for checking continuity but also essential for understanding derivative and integral calculus. In our problem, the limits at \(x = 3\) helped establish that the function is continuous everywhere.