Problem 78 - 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 \(\lim _{x \rightarrow 0^{+}} f(x)\). \(f(x)=x^{2} / e^{x} ; I=(-\infty, \infty)\)

Step-by-Step Solution

Answer

The graph of \(f(x) = \frac{x^2}{e^x}\) has a local maximum at \(x=2\) and a horizontal asymptote at \(y=0\).

Step 1: Determine Critical Points

To find the critical points of the function \(f(x) = \frac{x^2}{e^x}\), we first find the derivative: \(f'(x) = \frac{d}{dx}\left(\frac{x^2}{e^x}\right)\). Using the quotient rule, \(f'(x) = \frac{(e^x)(2x) - (x^2)(e^x)}{(e^x)^2} = \frac{2x - x^2}{e^x}\). Critical points occur when \(f'(x) = 0\) or is undefined. Setting \(f'(x) = 0\) gives \(2x - x^2 = 0\) which simplifies to \(x(x-2) = 0\). Therefore, the critical points are \(x = 0\) and \(x = 2\).

Step 2: Determine Concavity and Inflection Points

Find the second derivative to determine concavity: \(f''(x) = \frac{d}{dx}\left(\frac{2x - x^2}{e^x}\right)\). Applying the quotient rule again, \(f''(x) = \frac{(2 - 2x)(e^x) - (2x - x^2)(e^x)}{(e^x)^2} = \frac{2 - 4x + x^2}{e^x}\). Inflection points occur where \(f''(x) = 0\): set \(2 - 4x + x^2 = 0\). Solving, \(x = 2\) is the inflection point. This shows a sign change in concavity.

Step 3: Determine Horizontal Asymptote Using l'Hôpital's Rule

To find the horizontal asymptote as \(x \to \infty\), evaluate \(\lim_{x \to \infty} \frac{x^2}{e^x}\). This limit is in the indeterminate form of \(\frac{\infty}{\infty}\), so apply l'Hôpital's Rule. Differentiate the numerator and denominator: \(\lim_{x \to \infty} \frac{2x}{e^x}\). Apply l'Hôpital's Rule again: \(\lim_{x \to \infty} \frac{2}{e^x} = 0\). Therefore, the horizontal asymptote is \(y = 0\).

Step 4: Sketch the Graph

Combine all the information: \(f(x)\) is defined for all \(x\) and has critical points at \(x=0\) and \(x=2\). It changes from increasing to decreasing at \(x=2\), implying a local maximum there. The inflection point at \(x=2\) indicates a change in concavity. The graph approaches the horizontal asymptote \(y=0\) as \(x \to \infty\). Sketch these features for the complete graph outline.

Key Concepts

Critical Points

In calculus, critical points are where the derivative of a function either equals zero or does not exist. These points are essential since they indicate where the function may have local maxima, minima, or points of inflection.
To find the critical points for the function \( f(x) = \frac{x^2}{e^x} \), we compute the derivative, using the quotient rule, to find \( f'(x) = \frac{2x - x^2}{e^x} \).

• Set the derivative \( f'(x) = 0 \) to find: \( 2x - x^2 = 0 \)
• This simplifies to \( x(x-2) = 0 \), providing critical points at \( x=0 \) and \( x=2 \).

Once these points are determined, you can use other methods to analyze the nature of these critical points, whether they are a local maximum, minimum, or neither.

Inflection Points

Inflection points occur where a function changes its concavity. This can be identified by finding where the second derivative changes sign.
First, find the second derivative of \( f(x) = \frac{x^2}{e^x} \). Using the quotient rule again, we get \( f''(x) = \frac{2 - 4x + x^2}{e^x} \).

• Set \( f''(x) = 0 \) which simplifies to \( 2 - 4x + x^2 = 0 \).
• Solve the quadratic equation to find inflection points.

For this function, \( x = 2 \) is an inflection point, indicating a change in the curve's concavity at this point. This means that the graph of \( f(x) \) changes its curvature direction at \( x = 2 \), possibly contributing to varied graph patterns, like transitioning from concave up to concave down or vice versa.

l'Hôpital's Rule

l'Hôpital's Rule is a valuable tool in calculus for determining limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
In applying this to \( f(x) = \frac{x^2}{e^x} \), we find \( \lim_{x \to \infty} \frac{x^2}{e^x} \).

• The limit is indeterminate initially, so by differentiating the numerator and denominator, we obtain: \( \lim_{x \to \infty} \frac{2x}{e^x} \).
• Still indeterminate, apply l'Hôpital's Rule again to get: \( \lim_{x \to \infty} \frac{2}{e^x} = 0 \).

The repeated application of l'Hôpital's Rule helps simplify and solve for these tricky limits. This demonstrates that the function \( f(x) \) approaches 0 as \( x \) becomes very large.

Horizontal Asymptote

A horizontal asymptote of a function is a horizontal line that the graph of the function approaches as \( x \) moves toward positive or negative infinity. It is important as it provides insight into the end behavior of the function.
For \( f(x) = \frac{x^2}{e^x} \), using l'Hôpital's Rule helped determine that the function approaches 0 as \( x \to \infty \). Thus, the horizontal asymptote is \( y=0 \).

• This indicates that, though the portion of \( f(x) \) may extend upwards, ultimately, the graph trends towards flattening out along \( y=0 \) as \( x \) increases indefinitely.
• The horizontal asymptote helps predict how \( f(x) \) behaves with large inputs, offering critical information for graph sketching and comprehending the function's limits.

Understanding horizontal asymptotes is crucial for grasping the full extent of a curve's long-term trending behavior.