Question

State whether the indicated function is continuous at \(3 .\) If it is not continuous, tell why. $$ f(x)=\left\\{\begin{array}{ll} -3 x+7 & \text { if } x \leq 3 \\ -2 & \text { if } x>3 \end{array}\right. $$

Short answer

Yes, the function is continuous at 3.

Step-by-step solution

Evaluate Left-Hand Limit

For a function to be continuous at a point, both the left-hand and right-hand limits as well as the function value at that point should be the same. Start by evaluating the left-hand limit of the function as \( x \) approaches 3 from the left, which means using the expression \(-3x+7\). Substitute \( x = 3 \) into \(-3x+7\), we get: \[-3(3) + 7 = -9 + 7 = -2.\] Thus, the left-hand limit as \( x \to 3^- \) is \(-2\).

Evaluate Right-Hand Limit

Next, evaluate the right-hand limit of the function as \( x \) approaches 3 from the right, which uses the expression \(-2\). Since \( x > 3 \) results in a constant value, the right-hand limit as \( x \to 3^+ \) is \(-2\).

Determine Function Value at 3

The next step is to find the actual function value at \( x = 3 \). Since the inequality \( x \leq 3 \) includes 3, we use the expression \(-3x+7\) to find \( f(3) \). Evaluate \(-3(3) + 7\), which is \(-9 + 7 = -2\). So, \( f(3) = -2 \).

Compare Limits and Function Value

Finally, check whether the left-hand limit, right-hand limit, and the function value at \( x = 3 \) are the same. The left-hand limit is \(-2\), the right-hand limit is \(-2\), and the function value \( f(3) \) is \(-2\). Since all values are the same, the function is continuous at \( x = 3 \).

Key concepts

Left-Hand Limit

Understanding the left-hand limit is critical for determining a function's behavior as it approaches a particular point from the left side. If a function has different expressions for different intervals, like in a piecewise function, finding this limit involves using the expression defined for values approaching from the left.

In this exercise, the function takes the form \(-3x + 7\) for \(x \leq 3\). To find the left-hand limit as \(x\) approaches 3, substitute 3 into this expression:

• Compute \(-3(3) + 7\).
• You get \(-9 + 7 = -2\).

This indicates the left-hand limit at \(x = 3\) is \(-2\). Consistency between limits and function values at a point helps in ascertaining continuity of functions.

Right-Hand Limit

The right-hand limit considers the function as it nears a specific point from the right side. For piecewise functions, it usually leverages a different expression when \(x\) is greater than the target point.

In our example, for \(x > 3\), the function is simply \(-2\), as shown in the step-by-step solution. Therefore, as \(x\) approaches 3 from the right, the value of the function doesn't change - it remains \(-2\).

This constant output simplifies the right-hand limit evaluation since:

• Right-hand limit at \(x = 3\) is \(-2\).

Both the right-hand and the left-hand limits give insight into the function's behavior around \(x = 3\), crucial for verifying continuity.

Function Evaluation

Function evaluation at a given point involves determining the function's value using its definition. For a piecewise function, select the expression corresponding to the interval that includes the point.

For \(f(x)\) at \(x = 3\), use the expression linked with \(x \leq 3\), which is \(-3x + 7\). Substitute 3 in this expression:

• Compute \(-3(3) + 7\), resulting in \(-9 + 7 = -2\).

Thus, the function value \(f(3)\) equals \(-2\). Comparing this with left-hand and right-hand limits verifies continuity when they match at the point.

Piecewise Function

A piecewise function is defined as a function that has different expressions for different parts of its domain. It is essential to practice identifying which expression to use based on the input value.

In this scenario, the function is defined with two expressions:

• For \(x \leq 3\):
  • \(-3x + 7\)
• For \(x > 3\):
  • \(-2\)

This setup means each part of the domain uses a specific rule for calculating the function's value. When verifying continuity, check both limit expressions align with the defined point's value. Agreement between both sides and the point evaluation confirms a continuous function at that location.