Problem 47 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ f(x)=x e^{x} $$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, -1)\) and increasing on \((-1, \infty)\).

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we first find the derivative of the function. The given function is \( f(x) = x e^x \). Using the product rule (\((uv)' = u'v + uv'\)), where \( u = x \) and \( v = e^x \), we have:\[ f'(x) = (x)' \, e^x + x \, (e^x)' = 1 \, e^x + x \, e^x = e^x + xe^x = e^x(1 + x) \].

Step 2: Set the Derivative Equal to Zero

To find critical points where the function switches from increasing to decreasing or vice versa, set the derivative equal to zero:\[ e^x(1 + x) = 0 \] Since \( e^x \) is always positive, the equation simplifies to:\[ 1 + x = 0 \].This yields \( x = -1 \) as the critical point.

Step 3: Test Intervals Around the Critical Point

With the critical point \( x = -1 \), test the intervals around this point. Choose test points in the intervals \((-\infty, -1)\) and \((-1, \infty)\).- For \( x < -1 \), choose \( x = -2 \). We find \( f'(-2) = e^{-2}(1 - 2) = -e^{-2}, \) which is negative.- For \( x > -1 \), choose \( x = 0 \). We find \( f'(0) = e^0(1 + 0) = 1, \) which is positive.

Step 4: Determine Intervals of Increase and Decrease

Since \( f'(x) < 0 \) when \( x < -1 \), the function is decreasing on \((-\infty, -1)\). Since \( f'(x) > 0 \) when \( x > -1 \), the function is increasing on \((-1, \infty)\). Thus, the function decreases on \((-\infty, -1)\) and increases on \((-1, \infty)\).

Key Concepts

Intervals of Increase and Decrease

In calculus, understanding where a function increases or decreases is essential. These intervals help describe the behavior of the function as it moves along the x-axis. To identify these areas, we rely on the derivative of the function. The derivative tells us the slope of the tangent line to the curve at any given point.

• If the derivative is positive over an interval, the function increases in that interval.
• If the derivative is negative, the function decreases.

By checking different values of x around critical points, we can see where the function switches behaviors from increasing to decreasing or vice versa.

Derivatives

Derivatives are a key tool in calculus used to study change and slope. They provide the rate at which a quantity changes. For a function like our example, finding the derivative is the first step in understanding its behavior.The process involves different rules based on the function's form. Here, we used the product rule because our function is a product of two smaller functions: \( f(x) = x e^x \).

Using the Product Rule

The product rule is applied when taking the derivative of two multiplied functions. It is expressed as:\[(uv)' = u'v + uv'\]So, for \( u = x \) and \( v = e^x \), we found:\[f'(x) = 1 \cdot e^x + x \cdot e^x = e^x + xe^x \]The derivative tells us that changes in \( x \) influence \( e^x(1 + x) \). This expression is what we analyze to determine increases and decreases.

Critical Points

Critical points in a function are where its derivative equals zero or is undefined. They are important because these points can indicate where the function changes its behavior—shifting from increasing to decreasing or vice-versa.For our function \( f(x) = x e^x \), setting the derivative to zero gives:\[e^x(1 + x) = 0\]Since \( e^x \) never zeroes out, we solve for the other part:\[1 + x = 0\] This gives us a critical point at \( x = -1 \).

Analyzing Around the Critical Point

To decide the nature of the critical point, we test intervals around it. Evaluating the sign of our derivative in these intervals shows whether the function is increasing or decreasing:

• When \( x < -1 \), say at \( x = -2 \), the derivative is negative, indicating a decrease.
• For \( x > -1 \), like at \( x = 0 \), the derivative becomes positive, indicating an increase.

Therefore, the critical point \( x = -1 \) acts as a dividing line between these intervals.