Problem 43 - Thomas Calculus

Question

For what value of \(a\) is $$f(x)=\left\\{\begin{array}{ll}{a^{2} x-2 a,} & {x \geq 2} \\ {12,} & {x<2}\end{array}\right.$$ continuous at every \(x ?\)

Step-by-Step Solution

Answer

The function is continuous for \(a = 3\) or \(a = -2\).

Step 1: Understanding Continuity

A function is continuous at a point if the left-hand limit, the right-hand limit, and the function value at that point all equal each other. So, for the function to be continuous at every point, especially at \(x = 2\) where the function definition changes, the left-hand limit \(\lim_{x \to 2^-} f(x)\), the right-hand limit \(\lim_{x \to 2^+} f(x)\), and \(f(2)\) must be equal.

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

The left-hand limit as \(x\) approaches 2 from the left is simply \(f(x) = 12\) for \(x<2\). Thus, \(\lim_{x \to 2^-} f(x) = 12\).

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

For \(x \geq 2\), \(f(x) = a^2 x - 2a\). To find the right-hand limit as \(x\) approaches 2 from the right, we substitute: \[ \lim_{x \to 2^+} (a^2 x - 2a) = a^2(2) - 2a = 2a^2 - 2a.\] This expression must be equal to the left-hand limit for continuity.

Step 4: Set the Limits Equal for Continuity

Set the right-hand limit equal to the left-hand limit to ensure continuity at \(x=2\): \[ 2a^2 - 2a = 12. \]

Step 5: Solve the Equation

Simplify and solve the equation for \(a\): \[ 2a^2 - 2a - 12 = 0. \]Divide the entire equation by 2 to simplify:\[ a^2 - a - 6 = 0. \]Now, factor the quadratic:\[ (a - 3)(a + 2) = 0. \]Thus, \(a = 3\) or \(a = -2\).

Step 6: Verify Continuity for Both Values of a

Substitute \(a = 3\) in the expression for the right-hand limit:\[ 2(3)^2 - 2(3) = 2 \times 9 - 6 = 18 - 6 = 12. \]This matches the left-hand limit, so \(a = 3\) ensures continuity.Substitute \(a = -2\):\[ 2(-2)^2 - 2(-2) = 2 \times 4 + 4 = 8 + 4 = 12. \]This also matches the left-hand limit, so \(a = -2\) ensures continuity.Both \(a = 3\) and \(a = -2\) maintain continuity of the function at \(x=2\).

Key Concepts

Limits in Calculus

Limits are a foundational concept in calculus, determining the behavior of functions as inputs approach certain points.
By assessing limits, we understand a function's performance as it nears a specific value. This is especially crucial where a function might behave differently just before or after a point.
Key aspects of limits include:

• **Left-Hand Limit**: Denoted as \( \lim_{x \to c^-} f(x) \), it shows how the function approaches a point from the left.
• **Right-Hand Limit**: Expressed as \( \lim_{x \to c^+} f(x) \), it describes the function's approach from the right.
• **Limit Exists**: A limit at a point exists if both left-hand and right-hand limits are equal.

In analyzing our piecewise function, we examine both left and right limits at a switch point \(x=2\) to ensure they tally, ensuring continuity. By setting these equal, we solve for unknowns that maintain a seamless function.
Understanding limits is a key precursor to grasping continuity in more complex scenarios.

Piecewise Functions

Piecewise functions are unique functions that define separate expressions over different intervals.
This means that instead of one formula governing the whole range, multiple formulas control different sections. In our exercise:

• For \(x \geq 2\), the function is \(f(x) = a^2 x - 2a\).
• For \(x < 2\), the function is always 12.

Analyzing these sections independently is crucial because they may not intuitively form a continuous path.
We must ensure smooth transitions at the boundaries, like at \(x = 2\) where the definition switches. Only when the sections connect seamlessly can we consider the entire function continuous.
Solving for values of \(a\) helps manage these transitions, aligning the output of both sections at critical points.

Continuity at a Point

Continuity at a point means there's no sudden jump or gap in the function's value at that point.
A continuous function will have the graph unbroken and flowing smoothly through any given point.
For a function \(f(x)\) to be continuous at a point \(c\):

• \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)\)
• \(\lim_{x \to c} f(x) = f(c)\)

In the problem, continuity is checked at the cutoff point \(x = 2\), where different function expressions meet.
For continuity:- The left-hand limit \(\lim_{x \to 2^-} f(x)\) equals 12.- The right-hand expression \(a^2(2) - 2a\) must also result in 12.Solving this ensures the function stays smooth at \(x = 2\), allowing us to find feasible \(a\) values.
In this case, both \(a = 3\) and \(a = -2\) ensure a no-gap transfer from left to right at the transition point.