Problem 41 - Calculus with Concepts in Calculus

Question

Let \(f(x)=1 /\left(1+x^{2}\right)\) and \(g(x)=e^{-x^{2} / 2}\). Plot the graphs of \(f\) and \(g\) on \([-2,2]\), and determine which graph is more concave near the point \((0,1)\). Then justify your answer by comparing \(f^{\prime \prime}(0)\) and \(g^{\prime \prime}(0)\).

Step-by-Step Solution

Answer

The function \(f(x)\) is more concave near \((0,1)\) since \(f''(0) = -2\) and \(g''(0) = -1\).

Step 1: Define the Functions

We are given two functions:- The function \(f(x) = \frac{1}{1+x^2}\).- The function \(g(x) = e^{-x^2/2}\).Our goal is to compare their concavity near the point \((0,1)\) on the interval \([-2,2]\).

Step 2: Find the Second Derivative of f(x)

To analyze concavity, we need the second derivative. For \(f(x) = \frac{1}{1+x^2}\), we first find the first derivative by applying the quotient rule:\[f'(x) = \frac{0 - (1)(2x)}{(1+x^2)^2} = \frac{-2x}{(1+x^2)^2}\]Next, find the second derivative by differentiating \(f'(x)\):\[f''(x) = \frac{d}{dx}\left(\frac{-2x}{(1+x^2)^2}\right)\]Applying the quotient rule again:\[\begin{align*}f''(x) &= \frac{(-2)(1+x^2)^2 - (-2x)(2)(1+x^2)(2x)}{(1+x^2)^4} \&= \frac{-2(1+x^2)^2 + 8x^2(1+x^2)}{(1+x^2)^4} \&= \frac{-2(1+2x^2+x^4) + 8x^2 + 8x^4}{(1+x^2)^4} \&= \frac{-2 -4x^2 -2x^4+8x^2+8x^4}{(1+x^2)^4} \&= \frac{-2 + 4x^2 + 6x^4}{(1+x^2)^4}\end{align*}\]

Step 3: Evaluate f''(0)

Substitute \(x = 0\) into \(f''(x)\):\[f''(0) = \frac{-2 + 4(0)^2 + 6(0)^4}{(1+0^2)^4} = \frac{-2}{1} = -2\]

Step 4: Find the Second Derivative of g(x)

For the function \(g(x) = e^{-x^2/2}\), use the chain rule to find the first derivative:\[g'(x) = e^{-x^2/2} \cdot \left(\frac{d}{dx}(-x^2/2)\right) = e^{-x^2/2}(-x)\]Thus,\[g'(x) = -xe^{-x^2/2}\]Now find the second derivative:\[g''(x) = \frac{d}{dx}(-xe^{-x^2/2})\]Using the product rule:\[\begin{align*}g''(x) &= -\frac{d}{dx}(x)e^{-x^2/2} + x\frac{d}{dx}(e^{-x^2/2}) \&= -e^{-x^2/2} - x(e^{-x^2/2})(-x) \&= -e^{-x^2/2} + x^2 e^{-x^2/2}\end{align*}\]

Step 5: Evaluate g''(0)

Substitute \(x = 0\) into \(g''(x)\):\[g''(0) = -e^{0} + 0^2 e^{0} = -1\]

Step 6: Compare Concavity at (0,1)

The concavity at a point is determined by the sign and magnitude of the second derivative. We have:- \(f''(0) = -2\)- \(g''(0) = -1\)Since \(-2 < -1\), \(f(x)\) is more concave downward near \(x=0\) compared to \(g(x)\).

Key Concepts

Second Derivative Test

The Second Derivative Test helps us determine concavity and decide which way a curve bends at a specific point. It involves looking at the second derivative of a function at a specific point.

• If the second derivative is positive, the graph is concave upward at that point, resembling a "u" shape.
• If the second derivative is negative, the graph is concave downward, forming an "n" shape.
• If the second derivative equals zero, the test is inconclusive, and the point might be an inflection point.

When analyzing functions like those in the exercise, you evaluate the second derivatives, such as \( f''(0) = -2 \) and \( g''(0) = -1 \), to determine which function is more concave near a particular point.

Quotient Rule in Calculus

The Quotient Rule is essential when differentiating functions expressed as a quotient, or fraction, of two other functions. If you have a function \( f(x) = \frac{u(x)}{v(x)} \), the Quotient Rule states that its derivative is:\[\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\]When working with \( f(x) = \frac{1}{1+x^2} \), the numerator \( u(x) \) is \( 1 \), and the denominator \( v(x) \) is \( 1+x^2 \). Applying the rule results in the first derivative:\[f'(x) = \frac{-(2x)}{(1+x^2)^2}\]This method is crucial for finding further derivatives or solving complex calculus problems.

Product Rule in Calculus

The Product Rule is handy for finding the derivative of a product of two functions. For functions \( u(x) \) and \( v(x) \), the derivative of their product is given by:\[\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]In our exercise, to derive \( g(x) = e^{-x^2/2} \), we use this rule:1. Break down the terms as \( u(x) = -x \) and \( v(x) = e^{-x^2/2} \).2. Apply the Product Rule to find the second derivative:\[g''(x) = -\left(e^{-x^2/2}\right) + x^2\left(e^{-x^2/2}\right)\]This rule assists in calculating derivatives of compound expressions efficiently.

Function Graphing

Graphing functions aids in visualizing their behavior over an interval. By plotting functions like \( f(x) = \frac{1}{1+x^2} \) and \( g(x) = e^{-x^2/2} \) over the interval \([-2, 2]\), you can see how they grow or shrink, which helps in identifying concavity.

• Use graphing software or tools for precise plotting.
• Spot points of concavity by looking for areas where the curve changes direction.
• Compare different graphs by noting the bend around specific points.

In this exercise, visual tools help clarify why \( f(x) \) is more concave downward near \( x=0 \) than \( g(x) \).

Derivative Calculation

Calculating derivatives involves various rules, such as the Quotient and Product Rules, which help find the slope of the tangent to a curve at any point.

• The first derivative tells you the rate of change. For instance, in \( f(x) = \frac{1}{1+x^2} \), \( f'(x) = \frac{-2x}{(1+x^2)^2} \) describes how \( f(x) \) changes as \( x \) varies.
• The second derivative offers insight into concavity. Here, it's used to discover if a function is bending upwards or downwards at a point.

Through careful application of these rules, you can gain a deeper understanding of a function's behavior, as demonstrated in the exercise on \( f(x) \) and \( g(x) \).