Problem 66 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the values of \(a\) and \(b\) so that the following function is continuous everywhere. $$ f(x)= \begin{cases}x+1 & \text { if } x<1 \\ a x+b & \text { if } 1 \leq x<2 \\\ 3 x & \text { if } x \geq 2\end{cases} $$

Step-by-Step Solution

Answer

\(a = 4\) and \(b = -2\) make the function continuous everywhere.

Step 1: Understanding Continuity

For a function to be continuous at a point, it must be defined there, and its left-hand and right-hand limits must be equal to the function's value at that point. We will ensure this condition holds at the points where the function's definition changes, specifically at the boundaries \(x = 1\) and \(x = 2\).

Step 2: Checking Continuity at \(x = 1\)

Calculate the left-hand limit \(\lim_{{x \to 1^-}} f(x) = 1 + 1 = 2\). For the right-hand limit, use \(y = ax + b\): \(\lim_{{x \to 1^+}} ax + b = a \times 1 + b = a + b\). For continuity at \(x = 1\),\[2 = a + b\].

Step 3: Checking Continuity at \(x = 2\)

Calculate the left-hand limit again using \(y = ax + b\): \(\lim_{{x \to 2^-}} ax + b = a \times 2 + b = 2a + b\). Calculate the right-hand limit \(\lim_{{x=2^+}} f(x) = 3 \times 2 = 6\). For continuity at \(x = 2\),\[2a + b = 6\].

Step 4: Solving the System of Equations

We have two equations now: \(a + b = 2\) and \(2a + b = 6\). Subtract the first equation from the second to find the value of \(a\): \(2a + b - a - b = 6 - 2\), which simplifies to \(a = 4\). Substitute this back into the first equation to find \(b\): \(4 + b = 2\), resulting in \(b = -2\).

Step 5: Conclusion: Values of \(a\) and \(b\)

The values that make the function continuous everywhere are \(a = 4\) and \(b = -2\). Substitute these values back into the function segments to ensure each segment meets the continuity requirements at the specified points.

Key Concepts

Limits and Continuity

Understanding the continuity of a piecewise function is crucial in calculus. Continuity at a point means you can draw the function graph at that point without lifting your pencil. A function is continuous at a point if three conditions are met:

• The function is defined at that point.
• Its left-hand limit at that point exists and equals the function's value at that point.
• Its right-hand limit at that point exists and also equals the function's value at that point.

For piecewise functions, special attention is required at the points where the function's definition changes. In this exercise, we focus on the boundaries at \(x = 1\) and \(x = 2\). To verify continuity at these points, calculate:

• The left-hand limit by approaching the point from the left side.
• The right-hand limit by approaching from the right side.
• Ensure these limits match the function's value at that point for overall continuity.

Solving Systems of Equations

In order to make the function consistent and continuous across its entire range, we often need to solve systems of equations. These equations are derived from the continuity conditions as discussed. In this example, the function is designed to be continuous by adjusting parameters \(a\) and \(b\) for the linear components in the middle segment. After checking the continuity at the transition points, namely \(x = 1\) and \(x = 2\), two equations are result:

• \(a + b = 2\)
• \(2a + b = 6\)

This is a simple system of linear equations. By solving it step-by-step—first isolating one variable and substituting it back—you find the necessary values for \(a\) and \(b\). Such problems are an excellent practice for mastering equation systems and understanding how they tie into broader calculus concepts.

Calculus Problems

Calculus often involves solving problems related to limits, continuity, and differentiating or integrating functions. Piecewise functions, like the one in this exercise, appear often in calculus issues. They showcase real-world scenarios where behavior changes over intervals. When faced with problems involving piecewise functions and continuity, consider:

• Breaking down the function into its component parts.
• Applying calculus concepts such as limits to ensure smooth transitions at interval boundaries.
• Using algebra to solve for any unknowns that ensure the function behaves appropriately across its range.

Understanding these elements prepares you to tackle a range of calculus problems, from simple continuity checks to more complex integrations and derivations across non-uniform functions.