Problem 24 - Calculus

Question

find a value of \(k\) making \(h(x)\) continuous on [0,5] $$h(x)=\left\\{\begin{array}{ll} k x & 0 \leq x<1 \\ x+3 & 1 \leq x \leq 5 \end{array}\right.$$

Step-by-Step Solution

Answer

Set \(k = 4\) to make \(h(x)\) continuous.

Step 1: Understanding Continuity

A function is continuous on an interval if it is continuous at every point in that interval. For the given exercise, we need the function to be continuous at the point where the function's definition changes, which is at \(x = 1\). This means \( \lim_{{x \to 1^-}} h(x) = \lim_{{x \to 1^+}} h(x) = h(1) \).

Step 2: Calculate Left-Hand Limit

To find the left-hand limit as \(x\) approaches 1, evaluate \( \lim_{{x \to 1^-}} h(x) = k \cdot 1 = k \). This is because for values just less than 1, we use the piece defined as \(kx\).

Step 3: Calculate Right-Hand Limit and Function at x=1

To find the right-hand limit as \(x\) approaches 1, evaluate \(\lim_{{x \to 1^+}} h(x) = 1 + 3 = 4 \). This is because for values just greater than or equal to 1, we use the piece defined as \(x + 3\), and similarly, \(h(1) = 1 + 3 = 4\).

Step 4: Set Limits Equal for Continuity

For \(h(x)\) to be continuous at \(x = 1\), the left-hand limit must equal the right-hand limit and the function value at \(x=1\). Hence, set \(k = 4\). This makes both sides equal to 4, ensuring continuity at the boundary \(x = 1\).

Key Concepts

Piecewise Functions

Piecewise functions are mathematical expressions defined by different formulas or pieces corresponding to specific intervals in their domain. Each piece is applied based on the value of the independent variable, which in this case is \(x\). This makes piecewise functions quite versatile because they can describe scenarios where a rule changes at different points. In the given exercise, the function \(h(x)\) is defined as:

• \(kx\) for \(0 \leq x < 1\)
• \(x+3\) for \(1 \leq x \leq 5\)

Here, we have two distinct rules depending on the range of \(x\). The goal is to find a value of \(k\) that makes \(h(x)\) continuous over the entire interval \([0,5]\). The challenge usually arises at the point where the definition of the function changes, in this case, at \(x = 1\). Understanding how each piece operates on its interval is key to determining continuity.

Limits

In mathematical analysis, limits help us understand the behavior of a function as it approaches a specific point, even if the function is not explicitly defined at that point. When dealing with piecewise functions, limits are particularly important at the points where the definition of the function shifts from one piece to another.

When considering the function \(h(x)\) at \(x = 1\), we need to evaluate:

• The left-hand limit, \(\lim_{{x \to 1^-}} h(x)\), which for values of \(x\) just under 1, is represented by the piece \(kx\).
• The right-hand limit, \(\lim_{{x \to 1^+}} h(x)\), which for values of \(x\) equal to or just greater than 1, is represented by the piece \(x+3\).

To make the function continuous, we set these limits equal, ensuring they both converge to the actual function value at \(x = 1\). This approach helps us find the right value of \(k\) to maintain continuity at this point.

Continuity at a Point

Continuity at a point means that there are no gaps, jumps, or breaks in a function at that particular spot. For a function to be continuous at a certain point \(x = c\), three conditions must be satisfied:

• The function must be defined at \(x = c\), meaning \(h(c)\) exists.
• The limit of the function as \(x\) approaches \(c\) from the left (\(\lim_{{x \to c^-}} h(x)\)) must exist.
• The limit from the right (\(\lim_{{x \to c^+}} h(x)\)) must also exist.
• Both of these limits must equal the function value at \(x = c\).

In the exercise, the continuity at \(x = 1\) was established by equating the left-hand limit, right-hand limit, and the actual function value at \(x = 1\). Specifically, we found:- \(\lim_{{x \to 1^-}} h(x) = k \)- \(\lim_{{x \to 1^+}} h(x) = 4\)- And \(h(1) = 4\)Equating \(k = 4\) ensures these conditions match, thus making the function continuous at \(x = 1\), seamlessly blending both pieces of the function.