Problem 22 - Calculus for Biology and Medicine

Question

Determine all inflection points. $$ f(x)=x e^{-x}, x \geq 0 $$

Step-by-Step Solution

Answer

The inflection point is at \( x = 2 \).

Step 1: Find the First Derivative

To locate inflection points, we first need to find the second derivative. Start by finding the first derivative of the function. Let \( f(x) = x e^{-x} \). Use the product rule \( (uv)' = u'v + uv' \) to differentiate:\[ f'(x) = (1 \, \cdot \, e^{-x}) + (x \, \cdot \, (-e^{-x})) = e^{-x} - x e^{-x} = (1 - x) e^{-x} \]

Step 2: Find the Second Derivative

Differentiate the first derivative to get the second derivative. Differentiating \( f'(x) = (1-x) e^{-x} \):Apply the product rule again:\[ f''(x) = (-1)(e^{-x}) + (1-x)(-e^{-x}) = -e^{-x} - (1-x)e^{-x} \]Simplify:\[ f''(x) = -e^{-x} + (x-1)e^{-x} = (x-2)e^{-x} \]

Step 3: Solve for Points of Inflection

Inflection points occur when \( f''(x) = 0 \). Solve \( (x-2)e^{-x} = 0 \).Since \( e^{-x} \) is never zero, set \( x-2 = 0 \):\[ x-2 = 0 \Rightarrow x = 2 \]

Step 4: Verify the Inflection Point

To confirm \( x = 2 \) is an inflection point, check if the concavity changes around this point. Evaluate the second derivative on either side of \( x = 2 \).For \( x < 2 \) (e.g., \( x = 1 \)):\[ f''(1) = (1-2)e^{-1} = -e^{-1} < 0 \] - f is concave down.For \( x > 2 \) (e.g., \( x = 3 \)):\[ f''(3) = (3-2)e^{-3} = e^{-3} > 0 \] - f is concave up.There is a change in concavity, so \( x = 2 \) is indeed an inflection point.

Key Concepts

Calculus and Inflection Points

Calculus is a branch of mathematics that studies continuous change. It provides tools for understanding how functions behave, and one of these tools involves looking at **inflection points**.
Inflection points are where a curve changes its concavity. Simply put, it's the point where a curve goes from being cup-shaped (concave up) to being cap-shaped (concave down), or vice versa.

• Understanding where these points are situated helps in sketching the graph of a function correctly.
• They provide insight into the behavior of the function, particularly where the slope of its tangent changes direction.

Inflection points are not only critical in theoretical calculus but also useful in practical applications like physics, engineering, and economics, where they help in understanding rates of change and optimization problems.

Understanding the Second Derivative

The **second derivative** of a function offers crucial information about the function's concavity.The first derivative, denoted as \( f'(x) \), gives us the slope of the tangent line to the function at any given point.
But it's the second derivative, denoted as \( f''(x) \), that tells us how this slope itself is changing.

• If the second derivative is positive, \( f''(x) > 0 \), the function is concave up.
• If it's negative, \( f''(x) < 0 \), the function is concave down.
• If it equals zero, \( f''(x) = 0 \), it may indicate an inflection point, provided there is a sign change around that point.

For the function \( f(x) = x e^{-x} \), understanding the sign of \( f''(x) \) helps confirm whether \( x = 2 \) is truly an inflection point, which is critical in understanding how the graph behaves at and around that point.

Concavity Change and Inflection Point Verification

To verify an **inflection point**, it's essential that there is a change in **concavity** around that point.The second derivative test is the perfect tool for this verification process.
When analyzing \( f''(x) \), particularly for the function \( f(x) = x e^{-x} \), you need to evaluate the second derivative for values just before and after the potential inflection point.

• At \( x < 2 \), the second derivative \( f''(x) \) was found to be negative, indicating concave down.
• At \( x > 2 \), it was positive, indicating concave up.

This change in sign from negative to positive confirms that \( x = 2 \) is indeed an inflection point, as the concavity of the function changes there.
Identifying these changes gives a more complete understanding of the graph and the behavior of the function, which is exactly what calculus allows us to discern easily.