Problem 25 - 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 \leq 1 \\ 2 k x+3 & 1

Step-by-Step Solution

Answer

The value of \(k\) that makes \(h(x)\) continuous on \([0,5]\) is \(k = -3\).

Step 1: Understanding Continuity at x = 1

For the function \(h(x)\) to be continuous on the interval \([0,5]\), the left-hand and right-hand limits at \(x=1\) must be equal, and the function value \(h(1)\) must also equal these limits. Thus, we set \(h(1^-)=h(1^+)=h(1)\).

Step 2: Calculate Left-Hand Limit

For \(h(1^-)\), we use the expression \(kx\) from the first piece of the function. Evaluating \(kx\) at \(x=1\) gives \(h(1^-)=k \cdot 1 = k\).

Step 3: Calculate Right-Hand Limit

For \(h(1^+)\), we use the expression \(2kx + 3\) from the second piece of the function. Evaluating \(2kx + 3\) at \(x=1\) gives \(h(1^+)=2k \cdot 1 + 3 = 2k + 3\).

Step 4: Determine the Continuity Condition

Set the left-hand limit equal to the right-hand limit for continuity at \(x=1\). Thus, \(k = 2k + 3\).

Step 5: Solve for k

Rearrange the equation \(k = 2k + 3\) to isolate \(k\). This gives \(k - 2k = 3\) or \(-k = 3\). Solving for \(k\) gives \(k = -3\).

Key Concepts

Piecewise Functions

Piecewise functions are unique as they define different expressions over various intervals. In simpler terms, they can "switch" behaviors at different parts of their domain. This makes them very handy in modeling real-world situations where a rule or a formula changes at a certain point. For the function given, \( h(x) \):

• When \( 0 \leq x \leq 1 \), the function behaves like \( kx \).
• When \( 1 < x \leq 5 \), the function follows the pattern \( 2kx + 3 \).

Understanding how a piecewise function operates is key, especially when analyzing their continuity across different intervals.

Left-Hand Limit

The concept of the left-hand limit is central to understanding how a function behaves as it approaches a particular point from the left. For the given piecewise function, the left-hand limit at \( x = 1 \) is derived from the expression defined for \( 0 \leq x \leq 1 \). This simply involves evaluating \( kx \) as \( x \) approaches \( 1 \) from the left. Thus:

• Calculate \( h(1^-) \): Substitute \( x = 1 \) in \( kx \), getting \( k \cdot 1 = k \). This is the left-hand limit.

Knowing how to find the left-hand limit allows us to measure how the function "looks" just before reaching a specific point.

Right-Hand Limit

Where the left-hand limit looks to the left, the right-hand limit focuses on how a function behaves as it approaches a particular point from the right. For our function \( h(x) \), this limit is calculated from the second piece of the function, valid for \( 1 < x \leq 5 \). Hence:

• Calculate \( h(1^+) \): Substitute \( x = 1 \) in \( 2kx + 3 \), resulting in \( 2k \cdot 1 + 3 = 2k + 3 \). This is the right-hand limit.

Being able to determine the right-hand limit helps ensure we understand how the function will act after passing a specific point.

Solving Equations

Solving equations is the final step to ensuring continuity in piecewise functions. It involves setting the left and right-hand limits equal to each other and solving for the unknown. For the function to be continuous at \( x = 1 \):

• Set the limits equal: \( k = 2k + 3 \).
• Rearrange the equation by subtracting \( 2k \) from both sides: \( k - 2k = 3 \).
• Simplify: \( -k = 3 \).
• Finally, solve for \( k \) to find \( k = -3 \).

Solving such equations is vital in determining the specific values required for the function to remain uninterrupted or continuous over its entire domain.