Problem 76 - Calculus and its applications

Question

In Exercise \(73,\) let $$ p(x)=\left\\{\begin{array}{ll} 1.50 x, & \text { for } x \leq 20 \\ 1.25 x+k, & \text { for } x>20 \end{array}\right. $$ Find \(k\) such that the price function \(p\) is continuous at \(x=20\)

Step-by-Step Solution

Answer

The value of \(k\) is 5.

Step 1: Understand Continuity Condition

For a piecewise function to be continuous at a point, the two different expressions must equal each other at that point. This means for continuity at \(x=20\), we need: \(1.50 \times 20 = 1.25 \times 20 + k\).

Step 2: Calculate the Left Side Value

Calculate the value of \(1.50 \times 20\). This gives us the left side value of the continuity equation. Thus, \(1.50 \times 20 = 30\).

Step 3: Express the Right Side Equation

Substitute \(x = 20\) into the second expression: \(1.25 \times 20 + k\). Simplifying, we have \(25 + k\).

Step 4: Set the Two Expressions Equal

Set the left side equal to the simplified right side from Step 3: \(30 = 25 + k\).

Step 5: Solve for k

Isolate \(k\) by subtracting 25 from both sides of the equation: \(30 - 25 = k\). Therefore, \(k = 5\).

Key Concepts

Continuity

The concept of continuity in mathematics is essential when analyzing functions. A function is continuous at a point if there is no break, jump, or hole at that specific location. For a function like a piecewise function, continuity at a point means both the left-hand limit and right-hand limit, as well as the function's value at that point, must be equal. This ensures a smooth transition between the different parts or branches of the function.
In Exercise 73, we were tasked with ensuring the function was continuous at the point where the piecewise expressions met, specifically at \( x = 20 \). This entailed setting the values from both expressions equal at \( x = 20 \) to satisfy the continuity condition.

Piecewise functions

Piecewise functions are an important concept in Calculus that involve a function defined by different expressions depending on the region of the domain. These functions allow for flexible modeling of real-world scenarios where different rules apply over different intervals.
In our example, the function \( p(x) \) switches from \( 1.50x \) for \( x \leq 20 \) to \( 1.25x + k \) for \( x > 20 \). This means that for any value of \( x \) up to 20, the price is calculated using the first expression, while a different calculation applies when \( x \) exceeds 20.

• Piecewise functions can represent varying conditions, like pricing based on quantity purchased.
• The challenge often lies in ensuring continuity at the transition point by carefully determining constants, such as \( k \), in the function's definition.

Solving equations

Solving equations is a fundamental part of calculus and mathematical analysis. It involves finding the value (or values) that satisfy a given equation. In this exercise, the equation arose from the requirement of continuity at a certain point.
To find the constant \( k \), we used the continuity condition that \( 1.50 \times 20 = 1.25 \times 20 + k \). This effectively reduced to solving the linear equation \( 30 = 25 + k \). By isolating \( k \), subtracting 25 from both sides, we found \( k = 5 \). Solving the equation is essential to ensure the function meets the required mathematical and contextual conditions.

Mathematical analysis

Mathematical analysis is a broad field examining the underlying principles and foundations of calculus and real analysis. This includes understanding limits, continuity, and different types of convergence. In the case of Exercise 73, mathematical analysis comes into play to explore how a piecewise function creates a continuous whole.
We examine the transitions between different function rules and use mathematical techniques to ensure these transitions do not disrupt the function's continuity. Through analysis techniques like examining limits and solving for unknowns, we gain deeper insights into the behavior of functions and can apply these findings to practical scenarios. Understanding these principles is crucial for deep comprehension of continuous functions and their applications throughout calculus.

• Analysis helps bridge calculus concepts with real-world problems, ensuring solutions are both mathematically valid and contextually applicable.
• The fundamental aspects of mathematical analysis, like continuity and limit evaluation, are pivotal in understanding and modeling functional behavior.