Problem 14 - Applied Calculus

Question

Find the critical points of the function and classify them as local maxima or local minima or neither. $$g(x)=x e^{-3 x}$$

Step-by-Step Solution

Answer

The critical point is \(x = \frac{1}{3}\), and it is a local maximum.

Step 1: Find the derivative

First, differentiate the function \(g(x) = x e^{-3x}\) to find the critical points. Use the product rule, which states that if \( u(x) = x \) and \( v(x) = e^{-3x} \), then \( g'(x) = u'(x)v(x) + u(x)v'(x) \). Here, \( u'(x) = 1 \) and \( v'(x) = -3e^{-3x} \). So, we have:\[ g'(x) = (1)(e^{-3x}) + x(-3e^{-3x}) = e^{-3x} - 3xe^{-3x} \]This simplifies to:\[ g'(x) = e^{-3x}(1 - 3x) \]

Step 2: Solve for critical points

To find the critical points, set the derivative equal to zero:\[ e^{-3x}(1 - 3x) = 0 \]Since \(e^{-3x} eq 0\) for any value of \(x\), we only need to solve:\[ 1 - 3x = 0 \]Solving this gives:\[ 3x = 1 \]\[ x = \frac{1}{3} \]

Step 3: Classify the critical points

To classify the critical points, we need to evaluate the second derivative \(g''(x)\). Differentiate \(g'(x) = e^{-3x}(1 - 3x)\) using the product rule:First, differentiate \( e^{-3x}\) to get \( -3e^{-3x} \) and \(1 - 3x\) to get \(-3\).\[ g''(x) = -3e^{-3x}(1 - 3x) + e^{-3x}(-3) \]\[ g''(x) = -3e^{-3x} + 9xe^{-3x} - 3e^{-3x} \]\[ g''(x) = e^{-3x}(-6 + 9x) \]Evaluate \(g''(x)\) at \(x = \frac{1}{3}\):\[ g''\left(\frac{1}{3}\right) = e^{-1}(9\left(\frac{1}{3}\right) - 6) = e^{-1}(3 - 6) = -3e^{-1} \]Since \(g''\left(\frac{1}{3}\right) < 0\), \(x = \frac{1}{3}\) is a local maximum.

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of a product of two functions. If you have two functions, let's call them \( u(x) \) and \( v(x) \), their derivative is not simply the product of their derivatives. Instead, the product rule is given by:

• \( (uv)' = u'v + uv' \)

This means you take the derivative of the first function \( u \), multiply it by the second function \( v \) without its derivative, and then add it to the first function \( u \) without its derivative multiplied by the derivative of the second function \( v \).
In our problem, we use the product rule to differentiate \( g(x) = x e^{-3x} \):

• Here, \( u(x) = x \) and \( v(x) = e^{-3x} \).
• The derivative \( u'(x) = 1 \) and \( v'(x) = -3e^{-3x} \).

Plugging these into the product rule gives the derivative of the function, which is crucial for finding critical points.

Second Derivative Test

The second derivative test is a technique used to determine whether a critical point of a function is a local maximum, local minimum, or neither. To apply this test, you need to find the second derivative of the function. Once you have the critical points from the first derivative, evaluate the second derivative at these points to classify them.The test result is interpreted as follows:

• If the second derivative is positive at a critical point, the function has a local minimum at that point.
• If the second derivative is negative, the function has a local maximum at that point.
• If the second derivative is zero, the test is inconclusive.

In our exercise, after computing the second derivative \( g''(x) \) for the function \( g(x) = x e^{-3x} \), we substituted the critical point \( x = \frac{1}{3} \):

• We found \( g''\left(\frac{1}{3}\right) < 0\), indicating a local maximum at this point.

Hence, the second derivative test is a powerful method in identifying and verifying the nature of critical points.

Local Maxima and Minima

Local maxima and minima are important features of the graph of a function. These points are where the function reaches a highest or lowest value within a small surrounding neighborhood. To find local maxima or minima, you first need to identify the critical points, where the first derivative of the function equals zero.In this problem:

• We used the first derivative \( g'(x) \) to locate where the slope of the tangent line is zero. This led us to the critical point \( x = \frac{1}{3} \).

Once you have the critical points, utilizing the second derivative test helps classify them. Our calculation showed that at \( x = \frac{1}{3} \), \( g''\left(\frac{1}{3}\right) < 0 \), confirming a local maximum. Thus, understanding how to identify and verify local extrema is essential in analyzing functions.

Differentiation Techniques

Differentiation techniques are a set of rules or methods used to find the derivative of a function. Differentiation is key when analyzing and understanding the behavior of functions, particularly in finding rates of change, tangents, and critical points.Some common differentiation techniques include:

• Power Rule: For a function \( f(x) = x^n \), the derivative is \( nx^{n-1} \).
• Product Rule: As discussed, it is used for the derivative of products of two functions.
• Quotient Rule: Used when differentiating the division of two functions.
• Chain Rule: Used for differentiating compositions of functions.

In our exercise, we primarily focused on the product rule, since the function \( g(x) = x e^{-3x} \) is a product of \( x \) and \( e^{-3x} \). Knowing and properly applying these techniques is essential in solving calculus problems effectively.