Problem 40 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The function is not differentiable at \( x=1 \) and \( x=-1 \), but it is differentiable at \( x=0 \).

Step 1: Evaluate the function at the given points

In this step, we first need to evaluate the function \( g(x) = (x^2 - 1)^{2/3} \) at the specific points provided.\( g(1) = ((1)^2 - 1)^{2/3} = 0^{2/3} = 0\).\( g(0) = ((0)^2 - 1)^{2/3} = (-1)^{2/3} \). Using the property \( (-1)^{2/3} = ((-1)^2)^{1/3} = 1^{1/3} = 1 \), we find \( g(0) = 1 \).\ Similarly, \( g(-1) = ((-1)^2 - 1)^{2/3} = (0)^{2/3} = 0 \). Thus, we have the points \((1, 0), (0, 1), (-1, 0)\).

Step 2: Analyze the behavior around each point

Next, let's analyze the behavior of the function as \( x \) approaches the points 1, 0, and -1:- Near \( x = 1 \) and \( x = -1 \), since \( g(x) = ((x^2 - 1)^{2/3}) \), the derivative behaves like \( g'(x) \sim (x-1)^{-1/3} \) or \( (x+1)^{-1/3} \) respectively which indicate cusps - points where the derivative is not defined.- Near \( x = 0 \), we find that the function is linear, as indicated by its continuity and smoothness in this region. The derivative is defined here, so the slope can be calculated straightforwardly.

Step 3: Examine differentiability

The function \( g(x) \) is differentiable where the derivative exists and is finite.- At \( x=1 \) and \( x=-1 \), the derivative is not defined due to the involvement of non-integer powers leading to vertical tangents or cusp-like behavior.- At \( x=0 \), \( g(x) \) is differentiable, indicating smooth behavior. This is because it does not involve any non-integer powered locations, hence its derivative does exist.

Step 4: Conclusion on differentiability at points

The function\( g(x) = (x^2 - 1)^{2/3} \) is not differentiable at the points \( x = 1 \) and \( x = -1 \) due to the presence of cusps, as evidenced by the undefined nature of the derivative. However, it is differentiable at \( x = 0 \) because the function behaves smoothly, with a defined derivative here.

Key Concepts

Cusp Points

Cusp points are specific locations on a graph where the slope of the curve suddenly changes direction. These points are recognized because the left-hand and right-hand derivatives do not match or are not defined.

When analyzing the differentiability of functions, cusp points are important because a function is not differentiable at these points. Differentiability means the function has a derivative at that point, which implies it's smooth and does not have sharp turns or breaks. However, at a cusp, the derivative does not exist.

• In the function \(g(x) = (x^2 - 1)^{2/3}\), cusp points occur at \(x = 1\) and \(x = -1\). This is due to the nature of the root and exponent, causing steep, sharp turns in the graph.
• These points illustrate that while the function itself is continuous at these points, due to the multiple possible tangent slopes around them, the function cannot be considered differentiable.
  

The function's behavior around these points resembles a cusp, where the graph's shape converges sharply without forming a definite tangent line.

Domain and Continuity

Understanding the domain and continuity of a function is fundamental to analyzing its behavior.
The domain of a function refers to all possible values that can be input into the function without causing undefined behavior, such as division by zero or square roots of negative numbers.

For the function \(g(x) = (x^2 - 1)^{2/3}\), all real numbers are included in its domain since \(x^2 - 1\) can take any real number as its result. However, the nature of the power \(^{2/3}\) demands careful consideration for differentiability rather than domain restrictions.

• Continuity, on the other hand, is the property that ensures the function does not have gaps, jumps, or breaks. For \(g(x)\), the function passes through all its values smoothly when moving along the x-axis.

While \(g(x)\) is continuous across its domain, continuity does not necessarily imply it is differentiable at every point. Thus, parts like \(x=1\) and \(x=-1\) highlight this distinction, being continuous yet not differentiable due to cusp points.

Behavior of Functions Around Points

Understanding how a function behaves around specific points gives insight into its overall shape and properties. When analyzing \(g(x) = (x^2 - 1)^{2/3}\), it's essential to observe how it behaves as you approach different critical points: \(x=1\), \(x=0\), and \(x=-1\).

• Near \(x=1\) and \(x=-1\), the function behaves erratically, similar to a point where the graph peaks and turns sharply. This indicates areas where derivatives \((g'(x))\) do not exist, contributing to the cusp-like formations. The steep slope near these points suggests expansion and contraction without continuity in direction.
• Conversely, around \(x=0\), \(g(x)\) exhibits a defined and smooth slope. Here, the derivative is straightforwardly obtained, shown by the function's linear behavior. This section highlights the clear transition in slope, indicating smooth continuity as one analyzes movement closer or away from zero.

By evaluating the behavior around these points, one distinguishes between simple continuity and points of non-differentiability, gaining clarity on what impacts a function's differentiability.