Problem 31 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=\frac{x}{\sqrt{x^{2}+1}} \end{equation}

Step-by-Step Solution

Answer

Inflection point at \((0,0)\); no local extremes.

Step 1: Find the first derivative

To identify critical points, find the first derivative of the function. Using the quotient rule, where if \( u = x \) and \( v = \sqrt{x^2 + 1} \), then the derivative \( y' = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \), we find:\[ y' = \frac{\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}}{x^2+1} = \frac{1}{(x^2+1)^{3/2}} \]

Step 2: Identify critical points

To find critical points, set the first derivative equal to zero and solve for \(x\). Here, solving \( \frac{1}{(x^2+1)^{3/2}} = 0 \) gives no solutions because the fraction does not equal zero. Check the endpoints or possible undefined points, but for this function, no critical points derive from setting the derivative to 0.

Step 3: Evaluate the second derivative

Next, find the second derivative to determine concavity and inflection points. Differentiate the first derivative:\[ y'' = \frac{d}{dx}\left(\frac{1}{(x^2+1)^{3/2}}\right) \]Using the chain and quotient rules:\[ y'' = -\frac{3x}{(x^2+1)^{5/2}} \]

Step 4: Identify inflection points

Inflection points occur where the second derivative changes sign. Set the second derivative \(-\frac{3x}{(x^2+1)^{5/2}} = 0\), which simplifies to \(x = 0\). Check the sign change in the second derivative around \(x = 0\) to confirm an inflection point.

Step 5: Analyze function at critical points and endpoints

Critical points at \(-\infty\), \(+\infty\) are not part of the real domain for deriving finite extrema, as they correspond to limits approaching. Thus, analyze \( y \) at highlighted points or as \( x \rightarrow \infty \) and \( x \rightarrow -\infty \) because no finite critical points were found.

Step 6: Graph the function

Finally, graph the function. Note that the inflection point is at \((0, 0)\) since substituting in the original equation gives \(y = 0\) at \(x = 0\). There are no local or absolute extreme points based on the derivative analysis. Sketching shows symmetry about the origin and asymptotic behavior as \(y\approxeq x\) when \(x\) is very large.

Key Concepts

Critical Points

Critical points are found where the first derivative of a function is zero or undefined. These points can indicate where a function might have local maxima or minima, or they might simply highlight key features of the graph. In this exercise, we examined the function \( y = \frac{x}{\sqrt{x^2 + 1}} \). After finding the first derivative, \( y' = \frac{1}{(x^2 + 1)^{3/2}} \), we realized it remains positive for all real \( x \).

Since the derivative is never zero, this function does not have critical points that correspond to local extremes. However, it is still crucial during analysis to consider endpoints and any bounds where the function might behave differently. Critical points play a significant role in understanding the behavior of functions at specific values and can help in sketching graphs accurately.

Derivatives

Derivatives are essential tools in calculus that represent the rate of change of a function concerning its variables. The first derivative provides insights into the slope of a function's tangent and can help identify whether a function is increasing or decreasing. For our function, \( y = \frac{x}{\sqrt{x^2 + 1}} \), we used the quotient rule to find the first derivative, \( y' = \frac{1}{(x^2 + 1)^{3/2}} \).

This derivative shows that as \( x \) changes, the slope of the function is perpetually positive, indicating it is always gently increasing. Calculating the second derivative, \( y'' = -\frac{3x}{(x^2 + 1)^{5/2}} \), helps us assess concavity and detect inflection points by studying where changes might occur in the function's curve. Understanding derivatives is crucial for interpreting changes in behavior throughout the function.

Inflection Points

Inflection points occur where a function changes its concavity, which can be observed by analyzing the second derivative's sign changes. For the function \( y = \frac{x}{\sqrt{x^2 + 1}} \), we calculated the second derivative as \( y'' = -\frac{3x}{(x^2 + 1)^{5/2}} \).

Setting the second derivative equal to zero, we found that \( x = 0 \) is a point of interest. By testing intervals around \( x = 0 \), we confirmed a sign change, indicating that \( x = 0 \) is indeed an inflection point. At this point, the graph of the function smoothly transitions from concaving upwards to concaving downwards as \( x \) increases. Inflection points provide valuable information about the curvature of a graph and are key in understanding the overall shape of the function.

Extreme Points

Extreme points refer to maximum or minimum values a function can attain on its domain. These can be either local, where values are higher or lower than neighboring points, or absolute, which are the overall highest or lowest values in the given range. In analyzing our function \( y = \frac{x}{\sqrt{x^2 + 1}} \), we found no local extrema.

The function's behavior is such that as \( x \) approaches infinity, the graph tends toward an asymptote, and there are no specific maximum or minimum values within a finite region. Instead, the graph is continuously increasing, verifying our findings from the derivative analysis. Extreme points are essential for understanding where a function might "peak" or "dip," providing insights into its most significant values.