Problem 42 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\sqrt[3]{x^{3}+1}$$

Step-by-Step Solution

Answer

The function has a stationary inflection point at \(x = 0\) and no local or absolute extremes.

Step 1: Finding Critical Points

To find critical points, we first need the derivative of the function \(y = \sqrt[3]{x^3 + 1}\). Since this can be rewritten as \(y = (x^3 + 1)^{1/3}\), apply the chain rule: \[ y' = \frac{1}{3}(x^3 + 1)^{-2/3} \times 3x^2 = x^2 (x^3 + 1)^{-2/3} \] Critical points occur when the derivative is zero or undefined. Since \(x^2 \geq 0\), \(y' = 0\) when \(x^2 = 0\), which is at \(x = 0\). The derivative is defined for all \(x\), so no undefined points contribute critical points.

Step 2: Finding Inflection Points

Inflection points occur where the concavity changes, which means finding the second derivative \(y''\). First, apply the product rule and chain rule to \(y' = x^2 (x^3 + 1)^{-2/3}\): 1. Let \(u = x^2\) and \(v = (x^3 + 1)^{-2/3}\). Thus, \(y' = uv\).2. \(u' = 2x\) and \(v' = -2x^2 (x^3 + 1)^{-5/3}\) (by the chain rule).The second derivative \(y'' = u'v + uv'\) becomes:\[ y'' = 2x (x^3 + 1)^{-2/3} - 2x^2 \cdot 2x^2 (x^3 + 1)^{-5/3} \]Set \(y'' = 0\) to find possible inflection points. Simplification yields \(x = 0\) as a potential inflection point since the terms can be zero for \(x = 0\). Check sign changes to confirm.

Step 3: Testing for Local and Absolute Extremes

We test the sign of \(y'\) around \(x = 0\) to check if it's a local extreme point. For \(x < 0\), \(y'\) is positive because both \(x^2>0\) and \((x^3+1)^{1/3} < 0\). Hence, \(y'\) becomes negative which indicates a local maximum.For \(x > 0\), \(y'\) is positive indicating increasing, and hence \(x = 0\) is not a local extremum. We check the endpoints of the domain, which here are unbounded. No absolute extremes are found without further boundary definition.

Step 4: Graph and Confirm Findings

Graph \(y = \sqrt[3]{x^3 + 1}\) to visualize behavior. The graph shows symmetry to the x-axis, no absolute extremes over full domain, with a stationary inflection point at \(x = 0\). This confirms changes in concavity witnessed by changes in the second derivative:- Symmetry about x-axis suggests point of inflection.

Key Concepts

Critical Points

Critical points of a function are the values of \( x \) where the first derivative, \( y' \), is either zero or undefined. These points can help identify locations of potential local maxima or minima in a function.
To find the critical points of the function \( y = \sqrt[3]{x^3+1} \), we first determine its derivative. By rewriting the function in a more manageable form as \( y = (x^3 + 1)^{1/3} \), applying the chain rule gives us:

• \( y' = \frac{1}{3}(x^3 + 1)^{-2/3} imes 3x^2 = x^2 (x^3 + 1)^{-2/3} \)

Critical points occur where \( y' = 0 \) or if the derivative is undefined. Here, since \( y' \) is never undefined, we only check where \( x^2 = 0 \), leading to the critical point at \( x = 0 \). This is the only candidate for a critical point where potential changes in the function's rate occur.

Inflection Points

Inflection points are locations on the graph where the concavity changes. These points are found by taking the second derivative of a function.
Using the previous result for the first derivative of our function \( y = (x^3 + 1)^{1/3} \), the second derivative is calculated by using both the product and chain rules. Letting \( u = x^2 \) and \( v = (x^3+1)^{-2/3} \), this gives us:

• \( y'' = 2x (x^3 + 1)^{-2/3} - 2x^2 \cdot 2x^2 (x^3 + 1)^{-5/3} \)

Setting this second derivative to zero or checking points where the sign changes can identify inflection points. In our analysis, \( x = 0 \) emerged as a location where the concavity potentially changes, as the expression yielded a zero value.

Derivative

The derivative is a fundamental tool in calculus, representing the rate of change of a function's output with respect to its input. For any function \( f(x) \), the derivative \( f'(x) \) provides a mathematical expression describing this instantaneous rate of change.
In this exercise, finding \( y' \), the first derivative of \( y = (x^3 + 1)^{1/3} \), was crucial for identifying critical points. By rewriting the original function and applying the chain rule, the process yielded \( y' = x^2 (x^3 + 1)^{-2/3} \).

The first derivative is central in assessing how the function grows or shrinks over different intervals. When \( y' = 0 \), you have stationary points that may represent flat areas, peaks, or troughs on the function's curve.

Concavity

Concavity refers to the direction a function curves at a particular interval. It's vital for understanding the function's graphical behavior and is determined by the second derivative \( y'' \).

• If \( y'' > 0 \), the function is concave up, resembling a U-shape.
• If \( y'' < 0 \), it is concave down, similar to an upside-down U.

In the given problem, checking the sign of \( y'' \) helps understand how the graph of \( y = (x^3 + 1)^{1/3} \) bends concerning the x-axis. A change in sign of \( y'' \) at \( x = 0 \) indicates that this is an inflection point.
For a complete picture, it’s helpful to sketch or visualize the function, demonstrating areas of increasing or decreasing curvature, therefore solidifying your understanding of concavity and its impact on the overall graph shape.