Problem 43 - Essential Calculus

Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ g(x)=(x) 3^{-x} $$

Step-by-Step Solution

Answer

The function \( g(x) = x \cdot 3^{-x} \) has a single critical number at \( x = \frac{1}{\ln(3)} \). This is the point of a local maximum, with the function increasing for \( x < \frac{1}{\ln(3)} \) and decreasing for \( x > \frac{1}{\ln(3)} \).

Step 1: Find the derivative of the function

Start by finding the derivative of \( g(x) \). This will involve the product rule and the chain rule since \( g(x) = x \cdot 3^{-x} \). This gives \[ g'(x) = 3^{-x} - x \cdot \ln(3) \cdot 3^{-x} \]

Step 2: Find the critical numbers

Set \( g'(x) = 0 \) or observe where \( g'(x) \) does not exist. From the derivative, it is clear that \( g'(x) \) always exists. Solving \( g'(x) = 0 \) gives \[ 3^{-x} - x \cdot \ln(3) \cdot 3^{-x} = 0 \]\[ x = \frac{1}{\ln(3)} \] This is the only critical number.

Step 3: Find intervals where function is increasing or decreasing

Test intervals before and after the critical number in the derivative function:For \( x < \frac{1}{\ln(3)} \), \( g'(x) > 0 \), hence the function is increasing. For \( x > \frac{1}{\ln(3)} \), \( g'(x) < 0 \), hence the function is decreasing.

Step 4: Identify the relative extrema

The function switches from increasing to decreasing at \( x = \frac{1}{\ln(3)} \). Therefore, there is a local maximum at this point.

Step 5: Confirm results with a graphing utility

Graph the function \( g(x) = x \cdot 3^{-x} \) and observe that the results from the previous steps confirm with the graph. It should demonstrate a local maximum at \( x = \frac{1}{\ln(3)} \), with the function increasing for \( x < \frac{1}{\ln(3)} \) and decreasing for \( x > \frac{1}{\ln(3)} \).

Key Concepts

Product Rule

When differentiating a function that is the product of two or more simpler functions, you use the Product Rule. It's a powerful tool in calculus that simplifies finding derivatives of complex expressions. For a function defined as the product of two functions, say \( u(x) \) and \( v(x) \), the Product Rule states:

• \( (uv)' = u'v + uv' \)

Breaking this down:

• Differentiate \( u \) while keeping \( v \) constant.
• Differentiate \( v \) while keeping \( u \) constant.
• Add these two results together.

In the given problem, \( g(x) = x \cdot 3^{-x} \), you identify \( u(x) = x \) and \( v(x) = 3^{-x} \). Applying the Product Rule is crucial to correctly finding the derivative, \( g'(x) \), in the next steps.

Chain Rule

The Chain Rule is essential when dealing with compositions of functions. If you have a function inside another function, such as \( v(x) = 3^{-x} \) in our example, you'll rely on the Chain Rule to differentiate effectively.

• For a function \( v(x) = f(g(x)) \), differentiate the outer function \( f \) and multiply by the derivative of the inner function \( g \).
• Symbolically, this is \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

For \( 3^{-x} \), you need to recognize it as an exponential function, \( e^{-x \ln 3} \), where the outer function is the exponential \( e^x \) and the inner function is \(-x \ln 3 \). Using the Chain Rule means taking the derivative of \( e^{-x \ln 3} \) and multiplying by the derivative of \(-x \ln 3 \), which leads to \(-\ln 3 \cdot 3^{-x} \). This critical insight allows finding the derivative \( g'(x) \) accurately.

Increasing and Decreasing Intervals

Finding increasing and decreasing intervals involves analyzing where the derivative of the function, \( g'(x) \), is positive or negative.

• If \( g'(x) > 0 \), the function is increasing on that interval.
• If \( g'(x) < 0 \), the function is decreasing on that interval.

In this exercise, you first locate the critical number by setting \( g'(x) = 0 \). This gives the point \( x = \frac{1}{\ln(3)} \). To check the intervals, evaluate \( g'(x) \) on either side of this critical point.

• For \( x < \frac{1}{\ln(3)} \), you find \( g'(x) > 0 \), indicating the function is increasing.
• For \( x > \frac{1}{\ln(3)} \), \( g'(x) < 0 \), so the function is decreasing.

This method clearly shows where the function rises and falls.

Relative Extrema

Relative extrema refer to the local maximum or minimum values of a function. These occur at critical points where the function's derivative changes sign.

• A local maximum is where the function goes from increasing to decreasing.
• A local minimum is where the function goes from decreasing to increasing.

In the exercise with \( g(x) = x \cdot 3^{-x} \), since the function transitions from increasing to decreasing at \( x = \frac{1}{\ln(3)} \), this indicates a local maximum. It's confirmed by observing a sign change in \( g'(x) \). Using a graphing utility can help verify these conclusions visually, showing a peak at the identified critical point. This hands-on approach helps solidify understanding and confirms analytical results.