Problem 27 - Calculus

Question

find a value of \(k\) making \(h(x)\) continuous on [0,5] $$h(x)=\left\\{\begin{array}{ll} e^{k x} & 0 \leq x<2 \\ x+1 & 2 \leq x \leq 5 \end{array}\right.$$

Step-by-Step Solution

Answer

The value of \( k \) needed is \( k = \frac{\ln(3)}{2} \).

Step 1: Understand the Problem

The function \( h(x) \) is piecewise, meaning its definition changes at \( x = 2 \). To ensure \( h(x) \) is continuous on \([0, 5]\), the value of the function approaching \( x = 2 \) from both sides must be equal.

Step 2: Express Continuity at x = 2

The condition for continuity at \( x = 2 \) is that \( \lim_{x \to 2^-} h(x) = \lim_{x \to 2^+} h(x) = h(2) \). Thus, the exponential part must equal the linear part at \( x = 2 \).

Step 3: Calculate the Limit from the Left

As \( x \) approaches 2 from the left, \( h(x) = e^{kx} \). Therefore, \( \lim_{x \to 2^-} h(x) = e^{2k} \).

Step 4: Calculate the Limit from the Right

At \( x = 2 \), \( h(x) = x + 1 \), so both \( \lim_{x \to 2^+} h(x) = 3 \) and \( h(2) = 3 \).

Step 5: Set Limits Equal to Solve for k

For continuity at \( x = 2 \), set \( e^{2k} = 3 \). To find \( k \), solve the equation \( e^{2k} = 3 \).

Step 6: Solve for k

Take the natural logarithm of both sides: \( \ln(e^{2k}) = \ln(3) \). Simplifying, this gives \( 2k = \ln(3) \). Thus, \( k = \frac{\ln(3)}{2} \).

Key Concepts

Piecewise Functions

In mathematics, piecewise functions are defined by multiple sub-functions, each applying to a specific interval of the overall domain. This creates a contrast where each piece of the function is distinct over different parts of the domain. It is crucial in solving problems involving piecewise functions to recognize the specific intervals and ensure any constraints, like continuity, are respected.

• A function like \( h(x) \) mentioned in our problem takes different forms depending on the value of \( x \).
• For \( 0 \leq x < 2 \), an exponential form is used, \( e^{kx} \), while for \( 2 \leq x \leq 5 \), a linear component \( x+1 \) is applied.

Understanding where these changes occur (at our point of interest, \( x = 2 \)) is crucial not just for evaluating the function, but also ensuring properties like continuity remain intact.

Limits

The concept of limits is foundational in calculus, often used to define the behavior of a function as it approaches a particular point. Ensuring a function is continuous at a specific point involves matching the limits from both directions at that point.

• For a piecewise function, we take into consideration the left-hand limit \( \lim_{x \to c^-} \) and the right-hand limit \( \lim_{x \to c^+} \).
• In our example, it is necessary that \( \lim_{x \to 2^-} h(x) \) should equal \( \lim_{x \to 2^+} h(x) \) for continuity at \( x = 2 \).

By setting these limits equal, we can solve for any unknown variables (e.g., \( k \) in this exercise), ensuring that the function behaves predictably at transition points like \( x = 2 \).

Exponential Functions

Exponential functions, which involve the expression \( e^{kx} \), where \( e \) is Euler's number, represent rapid growth or decay depending on the sign and magnitude of \( k \). They are often used in continuous growth models and possess unique properties due to their constant percentage change over equal increments.

• In the context of piecewise functions, ensuring the exponential component aligns with other parts of the function often involves solving equations derived from limits and continuity properties.
• For the continuity example provided, setting \( e^{2k} = 3 \) allowed us to integrate the exponential and linear components of \( h(x) \).

Solving for \( k \) required natural logarithms, given the equation \( e^{2k} = 3 \), transforming into \( 2k = \ln(3) \) and finally \( k = \frac{\ln(3)}{2} \). This illustrates how the principles of exponential functions assist in ensuring a smooth transition between different parts of a piecewise function.