Problem 60 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the graph of the function \(f(x)=x^{2 / 3}\). Is the function continuous at \(x=0 ?\) Does \(f^{\prime}(0)\) exist? Why or why not?

Step-by-Step Solution

Answer

The function \(f(x) = x^{2/3}\) has a graph consisting of a horizontal parabola-like curve with all non-negative outputs. It is continuous at \(x=0\), since the limit as \(x\) approaches \(0\) equals \(f(0)\). However, the first derivative \(f'(x) = \frac{2}{3} x^{-1/3}\) does not exist at \(x=0\) due to the sharp corner on the graph, indicating that the function is not smooth at this point.

Step 1: Sketch the graph of \(f(x) = x^{2/3}\)

We can observe that the function \(f(x) = x^{2/3}\) includes all real numbers as its domain. Due to the exponent being a fraction, the output will always be non-negative for any real input value, \(x\). To sketch the graph, we can choose a couple of different values of \(x\) to get their corresponding \(y = f(x)\)-values: - When \(x = 0, y = f(0) = 0^{2/3} = 0\). - When \(x = 1, y = f(1) = 1^{2/3} = 1\). - When \(x = 8, y = f(8) = 8^{2/3} = 4\). - When \(x = -1, y = f(-1) = (-1)^{2/3} = 1\). - When \(x = -8, y = f(-8) = (-8)^{2/3} = 4\). Now we can plot these points and connect them to create the graph.

Step 2: Determine if the function is continuous at \(x = 0\)

A function is continuous at a point if the limit exists and is equal to the function's value at that point. We want to find the limit as \(x\) approaches \(0\) for the function \(f(x) = x^{2/3}\). We can see that \(f(0) = 0\), and as \(x\) approaches \(0\), the function's output also approaches \(0\) from both the left and the right sides. So, the function is continuous at \(x = 0\).

Step 3: Check if the first derivative exists at \(x = 0\)

We will find the first derivative of the function and then check if it is defined at \(x = 0\). To find the derivative using the power rule, we have: \( f'(x) = \frac{2}{3} x^{-1/3} \) Now we need to check if \(f'(0)\) exists: \( f'(0) = \frac{2}{3} (0)^{-1/3} \) The term \(0^{-1/3}\) is undefined, so the first derivative, \(f'(0)\), does not exist at \(x = 0\). This means that the function has a sharp corner at this point and is not smooth. In conclusion, the function \(f(x) = x^{2/3}\) is continuous at \(x = 0\), but its first derivative, \(f'(0)\), does not exist due to the sharp corner on the graph at this point.

Key Concepts

Continuity

Continuity is a fundamental concept in calculus that describes whether a function is smooth without any interruptions. A function is continuous at a point if you can draw its graph at that point without lifting your pencil. For a more formal definition, a function is continuous at some point \( x = a \) if the limit of the function as it approaches \( a \) from both the left and right matches the actual value of the function at \( a \). In mathematical terms, this is expressed as \( \lim_{{x \to a}} f(x) = f(a) \).
In the exercise above, the function \( f(x) = x^{2/3} \) was shown to be continuous at \( x = 0 \). This is because \( f(0) = 0 \), and as \( x \) approaches zero, \( f(x) \) also approaches zero. The graph of \( f(x) \) smoothly transitions at \( x = 0 \) without jumps or breaks, which is characteristic of continuity. While continuity is a telltale mark of a predictable function, it does not guarantee smooth derivatives, as seen with this function at zero.

Derivatives

Derivatives represent the rate at which a function's output changes with respect to changes in the input. Think of a derivative as a measure of how a function's slope evolves. Mathematically, taking the derivative is finding the function's slope at any given point, often denoted as \( f'(x) \).
The power rule is a basic tool for differentiation, especially useful with simple polynomials and rational exponents. It states that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). In our exercise with \( f(x) = x^{2/3} \), the derivative is \( f'(x) = \frac{2}{3}x^{-1/3} \).
Unfortunately, \( f'(0) \) does not exist because \( x^{-1/3} \) is undefined when \( x = 0 \). This result indicates a vertical tangent or a cusp at \( x = 0 \), breaking the smooth continuity derivatives typically suggest. So, while the function moves through zero continuously, it does so abruptly, making the derivative at zero impossible to define.

Graph Sketching

Graph sketching is a visual exploration to understand a function's behavior. By identifying key features such as intercepts, slopes, and curvature, students can gain insights into the function's overall shape and critical points.

• Identify and plot intercepts: For \( f(x) = x^{2/3} \), the critical intercept at \( x = 0 \) yields \( y = 0 \).
• Determine behavior around important points: Assess how the function behaves around \( x = 0 \), and other axis-crossing points.
• Use key values: Input different values to understand rising or falling trends. Sample values \( x = 1, x = 8, x = -1, x = -8 \) highlight the positive slope as \( y \) increases with \( x \).

The appearance of \( x^{2/3} \)'s graph indicates a soft, rounded peak at both positive and negative \( x \). While sketching, note the symmetry about the y-axis and the absence of drastic changes, supporting the continual nature of the function. As expressed in the exercise, this methodical approach reveals the discontinuities in derivatives more astutely when compared to raw calculations alone.