Problem 41 - Calculus Early Transcendentals

Question

For what value of the constant \(c\) is the function f continuous on \((-\infty, \infty) ?\) $$f(x)=\left\\{\begin{array}{ll}{c x^{2}+2 x} & {\text { if } x<2} \\\ {x^{3}-c x} & {\text { if } x \geqslant 2}\end{array}\right.$$

Step-by-Step Solution

Answer

The value of \(c\) is \(\frac{2}{3}\).

Step 1: Definition of Continuity

For a function to be continuous at a point \(x = a\), the following must hold true: 1. \(f(a)\) is defined.2. The limit \(\lim_{{x \to a}} f(x)\) exists.3. \(\lim_{{x \to a}} f(x) = f(a)\). In this exercise, we want \(f\) to be continuous everywhere on \(( -\infty, \infty)\). The primary point of concern is \(x = 2\) where the pieces of the function meet.

Step 2: Evaluate the Left-Hand Limit at x = 2

To find \(\lim_{{x \to 2^-}} f(x)\), use the expression for \(x < 2\). Thus, we have:\[ \lim_{{x \to 2^-}} (cx^2 + 2x) = c(2)^2 + 2(2) = 4c + 4. \]

Step 3: Evaluate the Right-Hand Limit and Value at x = 2

To find \(\lim_{{x \to 2^+}} f(x)\) and \(f(2)\), use the expression for \(x \geq 2\). Thus, we have:\[ f(2) = 2^3 - c(2) = 8 - 2c. \] Therefore, \( \lim_{{x \to 2^+}} f(x) = 8 - 2c. \)

Step 4: Set Limits Equal to Ensure Continuity

For the function to be continuous at \(x = 2\), the left-hand limit must equal the right-hand limit:\[ 4c + 4 = 8 - 2c. \]

Step 5: Solve for c

Solve the equation from Step 4:\[ 4c + 4 = 8 - 2c \]Add \(2c\) to both sides:\[ 6c + 4 = 8 \]Subtract 4 from both sides:\[ 6c = 4 \]Divide by 6:\[ c = \frac{2}{3} \]

Key Concepts

Piecewise Functions

A piecewise function is a function that is defined by different expressions based on the value of the input variable. These expressions apply to specific intervals of the input variable. In this exercise, the function \( f(x) \) is a piecewise function that uses two different expressions depending on whether \( x \) is less than 2 or greater than or equal to 2.

- When \( x < 2 \), we use the expression \( f(x) = cx^2 + 2x \).
- When \( x \geq 2 \), the expression becomes \( f(x) = x^3 - cx \).

Piecewise functions are common in real-world scenarios where different conditions or rules apply over different inputs. Understanding how to work with them is crucial in mathematics since many functions, especially in calculus, are defined this way.

Limits

In calculus, limits help us understand the behavior of functions as the input variable approaches a particular point. For piecewise functions, limits are especially important when approaching the boundary points where the expression changes.

To ensure a piecewise function is continuous at the boundary, the left-hand limit (approaching from the left) must equal the right-hand limit (approaching from the right). Here's how we calculated the limits at \(x = 2\) for the function in this exercise:
- Left-hand limit (\( x \to 2^- \)): We used \( cx^2 + 2x \) to find \( 4c + 4 \).
- Right-hand limit (\( x \to 2^+ \)): We used \( x^3 - cx \) to find \( 8 - 2c \).

Limits are the core of understanding continuity, making sure there's no gap or jump at the transition from one expression to another.

Calculus Problem-Solving

Solving calculus problems often involves several standard steps: identify the function type, check definitions, compute limits, and ensure all conditions are met for continuity or differentiability.

In this particular problem, to find the value of \(c \) that keeps the function continuous over all real numbers, we:
1. Identified it's a piecewise function with two parts.
2. Computed limits as \( x \to 2 \) from both sides, using each piece of the function.
3. Set the two limits equal to each other for continuity.
4. Solved the resulting equation for \( c \).

By carefully following these steps, we ensure all conditions needed for continuity are checked, leading us to a correct solution.

Definition of Continuity

Continuity is a fundamental concept in calculus. A function is continuous at a point if there is no interruption, gap, or jump at that point. According to the definition of continuity, a function \( f \) is continuous at \( x = a \) if:
- \( f(a) \) is defined.
- The limit \( \lim_{{x \to a}} f(x) \) exists.
- \( \lim_{{x \to a}} f(x) = f(a) \).

For this exercise, to ensure the function is continuous at \( x = 2 \), both the left-hand and right-hand limits must equal the function's value at that point. We calculated these limits separately and then solved for \( c \) by equating the two results. This thorough check ensures there is no "gap" or "jump" in the function, verifying its continuity across all real numbers.