Problem 26 - Calculus

Question

find a value of \(k\) making \(h(x)\) continuous on [0,5] $$h(x)=\left\\{\begin{array}{ll} k \sin x & 0 \leq x \leq \pi \\ x+4 & \pi

Step-by-Step Solution

Answer

No value of \(k\) makes \(h(x)\) continuous on [0, 5].

Step 1: Understand the Definition of Continuity

A function is continuous on an interval if it is continuous at every point within the interval. For a piecewise function to be continuous at a point where its definition changes, the limits approaching from both sides must be equal and equal to the function's value at that point.

Step 2: Identify the Point of Interest

The function switches definitions at the point where the first piece ends and the second begins, which is at \(x = \pi\). Therefore, we need to ensure continuity at \(x = \pi\).

Step 3: Apply Continuity Condition for \(x = \pi\)

For continuity at \(x = \pi\), we need that \(\lim_{x \to \pi^-} h(x) = h(\pi) = \lim_{x \to \pi^+} h(x)\).

Step 4: Calculate \(\lim_{x \to \pi^-} h(x)\)

As \(x\) approaches \(\pi\) from the left, we use the definition \(h(x) = k \sin x\). Thus, \(\lim_{x \to \pi^-} h(x) = k \sin \pi = k \cdot 0 = 0\).

Step 5: Calculate \(h(\pi)\)

Since \(h(x) = k \sin x\) at \(x = \pi\), \(h(\pi) = k \sin \pi = k \cdot 0 = 0\).

Step 6: Calculate \(\lim_{x \to \pi^+} h(x)\)

As \(x\) approaches \(\pi\) from the right, we use the definition \(h(x) = x + 4\). Thus, \(\lim_{x \to \pi^+} h(x) = \pi + 4\).

Step 7: Solve for \(k\) to Meet Continuity Condition

From Steps 4 to 6, we know \(0 = 0 = \pi + 4\). Therefore, \(\pi + 4 = 0\), which is impossible. So, the only way \(h(x)\) is continuous is if the segments don't meet.

Step 8: Conclude the Solution

The function cannot be made continuous on [0, 5] with any real value of \(k\) because there is no value that will satisfy the condition of continuity. So the problem statement might be incorrectly stated, or there is no solution.

Key Concepts

Piecewise Functions

Understanding piecewise functions is crucial in calculus. A piecewise function is defined by different expressions depending on the interval of the input. These functions are like patchwork quilts, where each patch represents a different expression.

In our original exercise, the function \( h(x) \) is a piecewise function with different expressions for different ranges of \( x \):

• \( k \sin x \) for \( 0 \leq x \leq \pi \)
• \( x+4 \) for \( \pi < x \leq 5 \)

Each expression applies over a specific range of \( x \), and where these ranges meet, ensuring continuity becomes essential. This is the point where the two pieces of the function must "seamlessly" connect.

To determine this seamless connection, we need to impose conditions that ensure the transition from one expression to another is smooth.

Limits

Limits help us understand how a function behaves as the input approaches a particular value. When dealing with piecewise functions, calculating limits at the points where the function changes its expression is vital.

In our exercise, the key point of interest is \( x = \pi \), where the function definition changes. When seeking continuity at this point, three key limits must be considered:

• \( \lim_{x \to \pi^-} h(x) \): This limit uses the piece of the function defined for \( 0 \leq x \leq \pi \).
• \( \lim_{x \to \pi^+} h(x) \): Using the expression valid for \( \pi < x \leq 5 \) provides this limit.
• \( h(\pi) \): The actual value of the function at \( x = \pi \) according to the corresponding piece.

Calculating these limits will tell us if the function approaches the same value from both the left and right as it approaches \( \pi \). If these limits match the value of the function at that point, the function is continuous here. However, if they don't match, it indicates a discontinuity.

Continuity Conditions

Continuity is when a function has no interruptions or jumps as it is graphed. For continuous piecewise functions, continuity conditions need to be satisfied at the points where the definition of the function switches.

For the function \( h(x) \) to be continuous at \( x = \pi \), the following condition must be met:

• \( \lim_{x \to \pi^-} h(x) = \lim_{x \to \pi^+} h(x) = h(\pi) \)

This condition ensures a smooth transition between the two function pieces.

In our exercise, both the limit as \( x \) approaches \( \pi \) from the left and the actual value at \( \pi \) is 0. However, the limit from the right is \( \pi + 4 \), resulting in a mismatch. This mismatch means that there is a jump or a gap right at \( x = \pi \). Thus, despite our efforts, no real value for \( k \) will fill this gap, meaning the function cannot be made continuous with any real value for \( k \) at this point.