Problem 48 - Calculus - AP Edition

Question

Show that $$f(x)=\left\\{\begin{array}{ll} x^{2}+2, & x \leq 1 \\ x+2, & x>1 \end{array}\right.$$ is continuous but not differentiable at \(x=1 .\) Sketch the graph of \(f\)

Step-by-Step Solution

Answer

The function is continuous at \(x=1\) but not differentiable there due to differing left and right derivatives.

Step 1: Understanding the Definition of Continuity

To show a function is continuous at a point, the function must satisfy three conditions at that point: (1) The function is defined at that point, (2) The limit of the function as it approaches the point from both sides exists, and (3) The limit matches the function's value at that point. We must check these conditions for the function \(f\) at \(x=1\).

Step 2: Checking the Function Value at x=1

Evaluate \(f(1)\). Since \(x=1\) falls under the first case \((x\leq 1)\), we use \(f(x) = x^2 + 2\). Substituting \(x=1\), we get \(f(1) = 1^2 + 2 = 3\). Thus, \(f(1) = 3\).

Step 3: Checking the Left-Hand Limit at x=1

Calculate \(\lim_{x \to 1^-} f(x)\) using \(f(x) = x^2 + 2\). The limit becomes \(\lim_{x \to 1^-} (x^2 + 2) = 1^2 + 2 = 3\).

Step 4: Checking the Right-Hand Limit at x=1

Calculate \(\lim_{x \to 1^+} f(x)\) using \(f(x) = x + 2\). The limit becomes \(\lim_{x \to 1^+} (x + 2) = 1 + 2 = 3\).

Step 5: Verifying Continuity at x=1

The function \(f\) is continuous at \(x=1\) because \(f(1) = 3\), \(\lim_{x \to 1^-} f(x) = 3\), and \(\lim_{x \to 1^+} f(x) = 3\). All conditions of continuity are satisfied.

Step 6: Understanding Differentiability

For \(f\) to be differentiable at a point \(x=a\), the derivative from the left must equal the derivative from the right at that point. The function must also be continuous, which we already verified.

Step 7: Calculating Left-Hand Derivative at x=1

The derivative of \(f(x) = x^2 + 2\) with respect to \(x\) is \(2x\). Thus, \(\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} 2x = 2\).

Step 8: Calculating Right-Hand Derivative at x=1

The derivative of \(f(x) = x + 2\) is \(1\). Thus, \(\lim_{x \to 1^+} f'(x) = 1\).

Step 9: Verifying Non-Differentiability at x=1

Since the left-hand derivative \(2\) is not equal to the right-hand derivative \(1\), the function is not differentiable at \(x=1\).

Step 10: Sketching the Graph

1. For \(x \leq 1\), plot the parabola \(y = x^2 + 2\). This segment is continuous up to \(x=1\) and curving upwards. 2. For \(x>1\), plot the line \(y = x + 2\). Starting smoothly from the point \((1, 3)\) with a straight line slope, distinct from the curve. The graph as a whole looks continuous, but the abrupt change in slope at \(x=1\) confirms the non-differentiability.

Key Concepts

Piecewise Functions

Piecewise functions are defined by different expressions based on different intervals in their domain. They allow us to describe situations where a function involves distinct formulas for different input values. In our given example, the function \( f(x) \) is defined by two separate expressions: for values of \( x \leq 1 \), it uses \( x^2 + 2 \), and for values of \( x > 1 \), it uses \( x + 2 \). To fully understand a piecewise function, it's important to consider:

• The different rules that apply to different intervals of \( x \).
• The points where the expression changes, known as the boundaries or endpoints, such as \( x = 1 \) in this case.
• Conditions such as continuity and differentiability at these points.

Piecewise functions are often used to accurately describe systems that react differently in different situations or domains, such as tax brackets, shipping costs, or—as we see here—different parts of a mathematical graph.

Limits and Derivatives

Understanding limits helps us grasp the behavior of piecewise functions at boundaries like \( x = 1 \). Limits tell us the value that a function approaches as we get closer to a specific point from either side. For continuity, the left-hand limit and right-hand limit need to match the actual value of the function at that point. In our exercise:

• The limit as \( x \) approaches 1 from the left is calculated using the expression \( x^2 + 2 \), resulting in the value 3.
• The limit from the right, using \( x + 2 \), also equals 3.
• Since both limits and the function's value at \( x = 1 \) match, the function is continuous at \( x = 1 \).

Derivatives, on the other hand, provide the slope or rate of change of a function at any given point. A function's differentiability requires that the derivatives from both sides at a point be equal. In this case:

• The left derivative, based on \( 2x \) from \( x^2 + 2 \), comes out to be 2 as \( x \) approaches 1.
• The right derivative, derived from \( f(x) = x + 2 \), is a constant 1.

With different derivative values at \( x = 1 \), we see the function is not differentiable here, indicating a sharp corner or cusp.

Graphing Functions

Graphing a piecewise function involves sketching each segment separately and ensuring they join correctly at the specified points. For our function \( f(x) \), the graph is a mix of a parabola and a straight line:

• For \( x \leq 1 \): We graph the parabola \( y = x^2 + 2 \), resulting in an upward curve up to the point \( (1, 3) \).
• For \( x > 1 \): We then draw the straight line \( y = x + 2 \), starting from the same point \( (1, 3) \) to maintain continuity.

The graph visually shows continuity at \( x = 1 \), where both segments meet seamlessly, but also highlights non-differentiability due to the sudden change in slope. The transition from a curving path to a straight line implies a shift in behavior, which is an insightful aspect common to piecewise functions and their graphs.