Problem 45 - Thomas Calculus

Question

For what values of \(a\) and \(b\) is $$f(x)=\left\\{\begin{array}{ll}{-2,} & {x \leq-1} \\ {a x-b,} & {-1 < x < 1} \\\ {3,} & {x \geq 1}\end{array}\right.$$ continuous at every \(x ?\)

Step-by-Step Solution

Answer

\(a = \frac{5}{2}\), \(b = -\frac{1}{2}\)

Step 1: Define Continuity

A function is continuous at a point if the limit from the left equals the limit from the right at that point and both equal the function value at that point. We need to check continuity at the points \( x = -1 \) and \( x = 1 \) where the piecewise definition changes.

Step 2: Check Continuity at \( x = -1 \)

First, evaluate the left-hand limit as \( x \) approaches \( -1 \): this is simply \( -2 \).Next, find the right-hand limit as \( x \) approaches \( -1 \), which involves the expression \( ax - b = a(-1) - b = -a - b \).For continuity at \( x = -1 \), set these equal to the function value at \( x = -1 \):\[-a - b = -2\].

Step 3: Check Continuity at \( x = 1 \)

First, evaluate the left-hand limit as \( x \) approaches \( 1 \): this is \( ax - b = a(1) - b = a - b \).Next, find the function value at \( x = 1 \), which is \( 3 \) (since \( x \geq 1 \) is covered).For continuity at \( x = 1 \), set the left limit equal to the function value:\[a - b = 3 \].

Step 4: Solve the System of Equations

We now have two equations:1. \(-a - b = -2\)2. \(a - b = 3\)Add these two equations to eliminate \( b \):\[-a - b + a - b = -2 + 3\]This simplifies to \(-2b = 1\), giving:\[b = -\frac{1}{2}\]Substitute \( b \) back into one of the original equations to find \( a \). Using the second equation:\[a - (-\frac{1}{2}) = 3\]\[a + \frac{1}{2} = 3\]\[a = 3 - \frac{1}{2} = \frac{5}{2}\].

Step 5: Verify Continuity

Substitute \( a = \frac{5}{2} \) and \( b = -\frac{1}{2} \) back into the expressions to verify they satisfy both conditions.For \( x = -1 \), the expression becomes:\[-a - b = -\frac{5}{2} + \frac{1}{2} = -2\].For \( x = 1 \), the expression becomes:\[a - b = \frac{5}{2} + \frac{1}{2} = 3\].Both are satisfied, confirming the calculations.

Key Concepts

Piecewise Functions

Piecewise functions are a special type of mathematical function, composed of multiple sub-functions, each defined on a specific interval or "piece" of the entire domain. This means that the function's expression changes depending on the value of the input, within certain bounds. In our example, the function has three different expressions:

• For \(x \leq -1\), the function \(f(x) = -2\).
• For \(-1 < x < 1\), it is defined as \(f(x) = ax - b\).
• For \(x \geq 1\), the function is \(f(x) = 3\).

Understanding piecewise functions is crucial because it allows us to analyze functions with different rules based on the input value. When working with such functions, it is important to ensure that all pieces are continuous, meaning there are no abrupt jumps at the points where the function switches from one piece to another.

Limits

Limits are foundational in calculus and are used to determine how a function behaves as it approaches a certain point. Limits help us understand the behavior of piecewise functions at the boundaries of each piece.

In a piecewise function, we assess the limit of each sub-function as it approaches the connecting points where the sub-functions meet, such as \(x = -1\) and \(x = 1\) in this case.

• The **left-hand limit** refers to the value that the function approaches as \(x\) approaches the boundary from the left side.
• The **right-hand limit** refers to the value the function approaches from the right side.

For the function to be continuous at these boundaries, the left-hand limit, right-hand limit, and the function's actual value at every boundary must be equal. This means there should be no gap or jump, ensuring smoothness. This definition becomes key when verifying the continuity of piecewise functions.

System of Equations

A system of equations is a set of two or more equations with common variables. Solving these systems means finding values for the variables that satisfy all the equations simultaneously. In the context of this exercise, after setting boundary conditions for continuity, we form a system of equations to solve for the unknowns, \(a\) and \(b\).

The conditions for continuity at \(x = -1\) and \(x = 1\) create:

• \(-a - b = -2\) from the point \(x = -1\)
• \(a - b = 3\) from the point \(x = 1\)

To solve this system, we can add the equations together to eliminate variable \(b\), allowing us to first find \(b\). Afterward, back-substitution gives us the value for \(a\). The solutions \(a = \frac{5}{2}\) and \(b = -\frac{1}{2}\) make the entire piecewise function continuous across all points.