Problem 8 - Calculus with Concepts in Calculus

Question

Find the intervals on which the graph of the function is concave upward and those on which it is concave downward. $$ g(x)=x^{2} e^{-x} $$

Step-by-Step Solution

Answer

The graph is concave upward on \((- fty, 1)\) and concave downward on \((1, \, fty)\).

Step 1: Find the First Derivative

To analyze concavity, we start by finding the first derivative of the function \( g(x) = x^2 e^{-x} \). Use the product rule: \( (uv)' = u'v + uv' \), where \( u = x^2 \) and \( v = e^{-x} \). Thus, the first derivative is: \( g'(x) = 2x e^{-x} + x^2 (-e^{-x}) = (2x - x^2)e^{-x} \).

Step 2: Find the Second Derivative

Now find the second derivative \( g''(x) \) to determine concavity. Differentiate \( g'(x) = (2x - x^2)e^{-x} \) using the product rule: \( g''(x) = [2 - 2x]e^{-x} + (2x - x^2)(-e^{-x}) \). Simplify to get: \( g''(x) = [2 - 4x + x^2]e^{-x} \).

Step 3: Solve for Inflection Points

To find inflection points, set \( g''(x) = 0 \): \( [2 - 4x + x^2]e^{-x} = 0 \). Since \( e^{-x} \) is never zero, solve \( 2 - 4x + x^2 = 0 \). Factor the quadratic equation to get \( (x-1)^2 = 0 \), giving a repeated solution at \( x = 1 \).

Step 4: Test Concavity Intervals

We need to test the intervals around the inflection point \( x = 1 \): Choose test points in the intervals \((-\infty, 1)\) and \((1, \, \infty)\). - For \( x = 0 \) in \((-fty, 1)\), \( g''(0) = 2e^{0} = 2 > 0 \), meaning concave upward. - For \( x = 2 \) in \((1, \, fty)\), \( g''(2) = [(2-8+4)e^{-2}] = -2e^{-2} < 0 \), meaning concave downward.

Key Concepts

Concavity

The concept of concavity is crucial in understanding the behavior of a graph in calculus. To put it simply, concavity describes how the graph of a function bends or "curves":

• If a graph is concave upward, it looks like an open bowl or "U" shape.
• If it is concave downward, it resembles an upside-down bowl or "n" shape.

The concavity of a function is closely linked to its second derivative. Specifically, if the second derivative of a function is positive over an interval, the graph is concave upward on that interval. Conversely, if the second derivative is negative, the graph is concave downward. Identifying the concavity of a function helps in predicting and sketching the general shape of its graph.

Derivatives

Derivatives are a fundamental tool in calculus that measure how a function changes. When finding the concavity of a function, both the first and the second derivatives come into play:

• First Derivative: It measures the rate of change of the function, giving insight into where the function is increasing or decreasing.
• Second Derivative: It provides additional information about the "curvature" or concavity of the function's graph.

In this exercise, we start by finding the first derivative of the function \( g(x) = x^2 e^{-x} \) using the product rule, which tells us about the slope of the graph.
Then, by finding the second derivative \( g''(x) \), we determine the shape of the graph, identifying intervals of concavity.
Understanding derivatives is key to analyzing the characteristics of a function's graph, including its concave nature.

Inflection Points

Inflection points are special points on the graph where the concavity changes. In other words, the graph switches from being concave upward to concave downward, or vice versa. Finding these points involves:

• Setting the second derivative \( g''(x) \) equal to zero, because these are potential points where the concavity may change.
• Solving the resulting equation to find the specific \( x \)-values which are candidates for inflection points.

In our problem, setting the second derivative to zero results in the equation \( 2 - 4x + x^2 = 0 \), leading us to a repeated solution at \( x = 1 \).
This value, \( x = 1 \), is our inflection point.
By testing intervals around this point, we confirmed that the graph changes concavity, which means \( x = 1 \) is indeed an inflection point. Understanding where inflection points occur tells us much about the overall shape and transitions in the graph.