Problem 81 - Calculus Single Variable

Question

A function \(f\) with domain either \(I=(-\infty, \infty)\) or \(I=(0, \infty)\) is given. Sketch the graph of \(f\). (The set \(C\) of critical points of \(f\) and the set \(I\) of inflection points of \(f\) are provided in some cases.) Use l'Hôpital's Rule to determine the horizontal asymptote of the graph. If \(I=(0, \infty),\) use l'Hôpital's Rule to determine \(\lim _{x \rightarrow 0^{+}} f(x)\). \(f(x)=x \ln (x) /\left(1+x^{2}\right) ; I=(0, \infty) ; C=\\{0.301 \ldots, 3.319 \ldots\\}\) \(I=\\{0.720 \ldots, 5.486 \ldots\)

Step-by-Step Solution

Answer

The horizontal asymptote is \(y = 0\). As \(x \to 0^+\), \(f(x) \to 0\).

Step 1: Identify the Domain

The domain of the function is given as \(I = (0, \infty)\). This means the function is only defined for positive values of \(x\).

Step 2: Find the Horizontal Asymptote Using l'Hôpital's Rule

To find the horizontal asymptote as \(x \to \infty\), evaluate \(\lim_{x \to \infty} \frac{x \ln(x)}{1+x^2}\). This is an indeterminate form \(\frac{\infty}{\infty}\). Using l'Hôpital's Rule, differentiate the numerator and denominator:- Derivative of numerator: \(\ln(x) + 1\)- Derivative of denominator: \(2x\)Apply l'Hôpital's Rule:\[ \lim_{x \to \infty} \frac{x \ln(x)}{1+x^2} = \lim_{x \to \infty} \frac{\ln(x) + 1}{2x} \]This simplifies to \(\frac{\ln(x)}{2x}\) whose limit as \(x \to \infty\) is \(0\). Thus, the horizontal asymptote is \(y = 0\).

Step 3: Find the Limit as \(x \to 0^+\)

Evaluating \(\lim_{x \to 0^+} \frac{x \ln(x)}{1+x^2}\), we encounter an indeterminate form \(\frac{0}{1}\). However, \(x \ln(x)\) as \(x \to 0^+\) tends to \(-\infty\), because \(\ln(x) \to -\infty\) while \(x \to 0^+\) is minimal.Thus, \(\lim_{x \to 0^+} x \ln(x) = 0\).Therefore, \(\lim_{x \to 0^+} \frac{x \ln(x)}{1+x^2} = 0\).

Step 4: Sketch the Function

Using the information obtained:- Horizontal asymptote at \(y = 0\).- As \(x \to 0^+\), \(f(x) \to 0\).- Critical points at around \(x = 0.301\) and \(x = 3.319\).- Inflection points at around \(x = 0.720\) and \(x = 5.486\).The graph starts near the origin, passes through the critical points where the slope changes, and approaches the asymptote as \(x\) increases.

Key Concepts

Function Sketching

To begin sketching your function, you need to gather as much information as possible about its behavior. Start with the domain of the function, given here as \((0,\infty)\). This means you are only focusing on positive \(x\) values. Knowing the domain helps set the boundaries for where your graph will be plotted.

Check for any asymptotes next. For horizontal asymptotes, the graph will approach but never quite reach a certain line as \(x\) grows very large. Critical points and inflection points further enrich your understanding by showing where the function changes direction or curvature.

When sketching, it helps to identify the behavior at the limits of \(x\). As \(x \to 0^+\), \(f(x)\) becomes very close to zero. The function also approaches a horizontal asymptote of \(y=0\) as \(x \to \infty\). These key features guide where you promptly adjust the curve of your sketch.

L'Hôpital's Rule

L'Hôpital's Rule is often used to resolve indeterminate forms like \(\frac{\infty}{\infty}\) or \(\frac{0}{0}\). This neat little trick involves taking the derivative of the function's numerator and denominator separately, then re-evaluating the limit.

In our exercise, we use L'Hôpital's Rule to find the horizontal asymptote. First, we differentiate the numerator \(x \ln(x)\), resulting in \(\ln(x) + 1\). Next, differentiate the denominator \(1 + x^2\) to get \(2x\).

Apply the rule, transform the limit expression into \(\lim_{x \to \infty} \frac{\ln(x) + 1}{2x}\), which eventually leads you to \(\lim_{x \to \infty} \frac{\ln(x)}{2x} = 0\). This confirms that the function's horizontal asymptote is indeed \(y = 0\).

Horizontal Asymptotes

Horizontal asymptotes direct how a function behaves at the extremities of its domain. In specific, these lines represent the value \(y\) that the function \(f(x)\) approaches as \(x\) either goes towards infinity or negative infinity.

In the current scenario, we evaluate the horizontal asymptote as \(x \to \infty\) using L'Hôpital’s Rule. Following through the calculus, you discover \(y = 0\) as the horizontal asymptote. Rapidly, this becomes a key feature when sketching the function, as it reveals how \(f(x)\) behaves for very large values of \(x\).

It's helpful to remember that not all functions will have horizontal asymptotes, but when they do, such lines provide bounds that the function approaches but never actually meets at large scales.

Critical Points

Critical points are places where the derivative of a function equals zero or becomes undefined. These points indicate potential maxima, minima, or points of inflection, where the function's direction might change.

For our function \(f(x) = \frac{x \ln(x)}{1+x^2}\), the provided critical points are around \(x = 0.301\) and \(x = 3.319\). Here, these are places where the slope of the function, which is its derivative, changes.

Understanding where these changes occur helps you predict the shape of the graph. By identifying where the function slopes upwards or downwards, you can patch together segments of the graph more precisely. Critical points serve as markers that divide distinct sections of the graph.

Inflection Points

An inflection point is where the curvature of the function changes, i.e., from being concave upwards to downwards, or vice versa. This is identified when the second derivative of a function crosses zero or becomes undefined.

In our example, inflection points are noted around \(x = 0.720\) and \(x = 5.486\). At these specific coordinates, you'll notice a change in the bending or curving of the graph. Inflection points can sometimes coincide with critical points but not always; they primarily alter the overall concavity rather than slope directly.

Recognizing inflection points can prevent misinterpretation of the graph's shape. As they serve critical roles in the flow of the curve, these points ensure you sketch a graph that accurately portrays how \(f(x)\) behaves, giving students clarity on its dynamic ups and downs.