Problem 19 - Calculus

Question

find \(k\) so that the function is continuous on any interval. $$f(x)=\left\\{\begin{array}{ll} k x & x \leq 3 \\ 5 & 3

Step-by-Step Solution

Answer

k = \frac{5}{3} ensures the function is continuous.

Step 1: Understand Continuity

For a piecewise function to be continuous on any interval, the left-hand limit and the right-hand limit at the point where the function changes, must be equal to each other and also equal to the value of the function at that point.

Step 2: Equation for Continuity at x = 3

Since the function changes its definition at \( x = 3 \), we set up the equation: \( \lim_{{x \to 3^-}} f(x) = \lim_{{x \to 3^+}} f(x) = f(3) \). This means \( 3k = 5 \), since the left-side function is \( kx \) and the right-side function is 5.

Step 3: Solve for k

From the equation \( 3k = 5 \), solve for \( k \) by dividing both sides by 3: \( k = \frac{5}{3} \).

Step 4: Verify the Solution

To verify, substitute \( k = \frac{5}{3} \) back into the function. The function is continuous at \( x = 3 \), as both \( 3\cdot\frac{5}{3} = 5 \) and \( 5 \) meet the condition for continuity.

Key Concepts

Understanding Limits in Continuity

In mathematics, a limit is a fundamental concept that helps us understand how a function behaves as it approaches a particular point. When dealing with piecewise functions, limits become essential for assessing continuity at the points where the definition of the function changes.

To ensure a function is continuous at a certain point, the left-hand limit (as you approach the point from the left), the right-hand limit (as you approach from the right), and the function's value at that point, all need to be the same.

For example, in the original exercise, the function changes definition at \( x = 3 \). For \( f(x) \) to be continuous at \( x = 3 \), the limit approaching from the left \( \lim_{{x \to 3^-}} f(x) \) must equal the limit approaching from the right \( \lim_{{x \to 3^+}} f(x) \), and both must equal \( f(3) \). This is a key step in checking the continuity of piecewise functions.

Exploring Piecewise Functions

Piecewise functions are unique because they are defined by different expressions based on different parts of the domain.

Understanding them involves recognizing where each piece applies and ensuring these pieces "fit" together in a way that doesn't break the function's path as we move along its domain. The function in the exercise transitions at \( x = 3 \), dividing into two expressions.

For values less than or equal to 3, the function is \( kx \). For values greater than 3, it is a constant, 5. When working with piecewise functions, especially in seeking continuity, you often will deal with equations that ensure the transition from one piece to the next is seamless.

• Identify all transition points where function definitions change.
• Assess continuity by ensuring at each transition point, limits must align.
• This typically means solving equations that set these limits equal to each other.

Solving Equations for Continuity

To solve an equation for continuity, especially in piecewise functions, we use the equality of limits principle.

First, you assemble an equation based on the requirement that at a transition point, like \( x = 3 \), the limits from both directions must equal the function's value at the point. In the problem given, this means setting \( 3k = 5 \) because as \( x \to 3^- \), the expression is \( kx \) and for \( x \to 3^+ \), it is 5.

Solve for \( k \) by rearranging and simplifying the equation. Divide both sides by 3 to isolate \( k \). Once you find \( k \), substitute back to verify that all conditions for continuity are fulfilled. This final check assures the accuracy and that the function behaves continuously through the specified point, ensuring it "flows" without jumps or breaks.