Problem 46 - Calculus: Early Transcendentals

Question

Zoom in toward the points \( (1, 0) \), \( (0, 1) \), and \( (-1, 0) \) on the graph of the function \( g(x) = (x^2 -1)^{2/3} \). What do you notice? Account for what you see in terms of the differentiability of \( g \).

Step-by-Step Solution

Answer

\( g(x) \) is non-differentiable at \( (1,0) \) and \( (-1,0) \), but differentiable at \( (0,1) \).

Step 1: Graph the Function

Begin by graphing the function \( g(x) = (x^2 - 1)^{2/3} \). The graph will show how \( g(x) \) behaves around the critical points \( (1,0) \), \( (0,1) \), and \( (-1,0) \). Pay attention to the shape and continuity of the graph as \( x \) approaches these points.

Step 2: Analyze the Point \((1, 0)\)

Zoom in on the point \((1, 0)\). Observe that as \( x \to 1 \), \( g(x) \) approaches 0 from both the left and right. Note the sharp corner at \( (1, 0) \), indicating a point where the derivative is not defined.

Step 3: Analyze the Point \((0, 1)\)

Focus on the point \((0, 1)\). As \( x \to 0 \), \( g(x) \) remains constant at 1. The function is smooth here, indicating differentiability because there is no sharp turn.

Step 4: Analyze the Point \((-1, 0)\)

Examine the graph near \((-1, 0)\). Like point \((1, 0)\), the function \( g(x) \) approaches 0. A similar sharp point is observed, which suggests that the derivative does not exist at \( x = -1 \).

Step 5: Conclusion on Differentiability

At points \((1,0)\) and \((-1,0)\), \( g(x) \) has cusps or corners, which indicate non-differentiability. At \((0, 1)\), the function is smooth, indicating differentiability.

Key Concepts

Cusps

Cusps are points on the graph where the function suddenly changes direction, creating a sharp corner. Imagine g(x) as a road and cusps as sudden hairpin turns. When you see points \(1,0\) and \(-1,0\) on the graph of \(g(x) = (x^2 - 1)^{2/3}\), the road doesn't smoothly curve; it makes a tight, abrupt turn. This sharpness indicates the presence of a cusp.
This is important because these cusps imply that the derivative, which measures the slope of the tangent to the curve, doesn’t exist at these points. Since you can't draw a single tangent line at a sharp corner, the function is non-differentiable here. Understanding cusps helps us know where and why a function isn’t differentiable.

Graph of Function

Visualizing the graph of a function is crucial for understanding its behavior and properties like differentiability. For the function \(g(x) = (x^2 -1)^{2/3}\), graphing it can show how it approaches key points.

• At \( (1, 0) \) and \( (-1, 0) \), the graph reveals the sharp points where differentiability is a problem.
• At \( (0, 1) \), the graph appears smooth, suggesting that at this point, the function is well-behaved and differentiable.

By graphing functions, you can visually detect where the function may be non-differentiable due to sharp corners or cusps. This allows you to predict and understand differentiability or lack thereof at certain points.

Critical Points

Critical points of a function are those where the behavior changes significantly. These could be places where the function reaches a local maximum, minimum, or where the derivative doesn’t exist.

• In our function \(g(x)\), the critical points \(1,0\) and \(-1,0\) are special because they contain cusps. This absence of a derivative at these points classifies them as critical.
• Conversely, \( (0, 1) \) remains differentiable and is not a problematic critical point concerning differentiability.

Analyzing critical points allows you not only to understand where a function changes in behavior but also helps identify places where it fails to be differentiable. This understanding is essential in calculus for optimizing and understanding function behavior.