Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\frac{e^{x}}{x^{2}+1} .\) Find and classify the critical points.

Step-by-Step Solution

Answer

The function \(f(x)=\frac{e^{x}}{x^{2}+1}\) has a critical point at \(x=1\). The point is neither a local maximum nor a local minimum as the second derivative test is inconclusive at \(x=1\).

Step 1: Find the derivative of the function

Applying the quotient rule of differentiation which states that \(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\), where \(u = e^x\) and \(v = x^2 + 1\). Thus, \(u' = e^x\) and \(v' = 2x\). Therefore, \[f'(x) = \frac{(x^2+1)e^x - 2x e^x }{(x^2 + 1)^2}\]

Step 2: Find the critical points

Now, set \(f'(x) = 0\) and solve for \(x\). This gives: \[0 = \frac{(x^2+1)e^x - 2x e^x }{(x^2 + 1)^2}\] Simplifying, the equation becomes: \[0 = (x^2 - 2x + 1)e^x\] So, \(x = 1\) is a critical point.

Step 3: Classify the critical points

Finally, classify the critical point. Take the second derivative of the function and substitute \(x =1\) into the second derivative, \(f''(x)\). The critical point is a local minimum if \(f''(1) >0\), it is a local maximum if \(f''(1)<0\) and the test is inconclusive if \(f''(1) =0\).

Step 4: Calculation of the second derivative

Finding the second derivative \(f''(x)\) requires applying the quotient rule twice. After some derivative calculations, we will have: \[f''(x)=\frac{(x^2-6x+1)e^x}{(x^2+1)^3}\]

Step 5: Classification of the critical point

Substitute \(x = 1\) into \(f''(x)\) to retrieve \(f''(1)=0\). Thus, the second derivative test is inconclusive.

Key Concepts

Understanding the Quotient Rule

When you need to differentiate a function that is expressed as a quotient of two functions, the quotient rule is your go-to tool. It makes the process straightforward by providing a specific formula:
\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\]Here's how it works:

• Identify the two parts of your function — the numerator (\(u\)) and the denominator (\(v\)).
• Differentiate both parts: \(u'\) is the derivative of the numerator and \(v'\) is the derivative of the denominator.
• Substitute into the formula to find the derivative of the quotient.

For the given function \(f(x) = \frac{e^x}{x^2 + 1}\), the numerator is \(e^x\) and the denominator is \(x^2 + 1\). By calculating \(u'\) as \(e^x\) and \(v'\) as \(2x\), you then apply these into the quotient rule formula.
This differentiation provides a useful expression for finding critical points, which are values of \(x\) where the derivative equals zero or is undefined.

Exploring the Second Derivative Test

The second derivative test determines if a critical point is a local maximum, minimum, or an inflection point. Here's a breakdown of how it works:

• Calculate the second derivative of the function \(f(x)\).
• Substitute the critical point found from the first derivative into the second derivative.
• Assess the nature of the critical point:
  • If \(f''(x) > 0\), it's a local minimum.
  • If \(f''(x) < 0\), it's a local maximum.
  • If \(f''(x) = 0\), the test is inconclusive, indicating you might need another method to determine the behavior.
    

In the exercise, when \(f''(1) = 0\), the second derivative test does not provide a classification, suggesting that further investigation is needed to understand the point's characteristic.

The Basics of Differentiation in Calculus

Differentiation is a fundamental concept in calculus, used to analyze the rate at which things change. At its core, differentiation provides the derivative, which can inform us of the slope of a tangent to a curve at any given point.
With functions like \(f(x) = \frac{e^x}{x^2 + 1}\), differentiation allows us to solve for critical points — spots on the curve where the function reaches local maximums or minimums, or possibly changes concavity.
To find the critical points:

• Compute the first derivative and set it to zero to locate critical points.
• Use the second derivative to determine the nature of these points, as shown earlier.

Differentiation, therefore, is essential not only for classifying critical points but also for understanding the broader behavior of functions in various mathematical and real-world contexts.