Problem 24 - Calculus for Biology and Medicine

Question

Determine all inflection points. \(f(x)=e^{-x^{2}}, x \geq 0\)

Step-by-Step Solution

Answer

The inflection point is at \(x = \frac{1}{\sqrt{2}}\).

Step 1: Find the first derivative

To determine potential inflection points, we first need to find the first derivative of the function. The original function is: \[ f(x) = e^{-x^2} \]The first derivative, using the chain rule, is:\[ f'(x) = -2xe^{-x^2} \]

Step 2: Find the second derivative

Next, we find the second derivative to locate inflection points, which occur when this derivative equals zero or is undefined. Differentiate the first derivative:\[ f'(x) = -2xe^{-x^2} \]Using the product rule, the second derivative is:\[ f''(x) = -2e^{-x^2} + 4x^2e^{-x^2} \]The simplified form is:\[ f''(x) = e^{-x^2}(-2 + 4x^2) \]

Step 3: Set the second derivative equal to zero

Find the values of \(x\) that make \(f''(x) = 0\). Setting the expression we found in the second derivative to zero:\[ e^{-x^2}(-2 + 4x^2) = 0 \]The factor \(e^{-x^2}\) is never zero, so we solve:\[ -2 + 4x^2 = 0 \]which simplifies to:\[ 4x^2 = 2 \]\[ x^2 = \frac{1}{2} \]\[ x = \frac{1}{\sqrt{2}} \]

Step 4: Verify each inflection point with a sign change

To ensure \(x = \frac{1}{\sqrt{2}}\) is a true inflection point, check the sign change of \(f''(x)\) around this point.- For \(x < \frac{1}{\sqrt{2}}\), pick \(x = 0\), then \(f''(0) = e^0(-2) = -2\), which is negative.- For \(x > \frac{1}{\sqrt{2}}\), pick \(x = 1\), then \(f''(1) = e^{-1}(-2 + 4) = 2e^{-1}\), which is positive.The change from negative to positive confirms an inflection point at \(x = \frac{1}{\sqrt{2}}\).

Step 5: Confirm valid domain restrictions

Since we are given \(x \geq 0\), there is no issue with \(x = 0\). Therefore, the inflection point found is within our domain.

Key Concepts

First Derivative

The first derivative of a function provides information about the function's rate of change. It helps us determine where the function is increasing or decreasing. To find the first derivative of the function \( f(x) = e^{-x^2} \), we apply the chain rule.

The chain rule suggests we differentiate the outer function and then multiply by the derivative of the inner function, \(-x^2\). Here's the process:

• Differentiate the outer function, \( e^u \), to get \( e^u \).
• Differentiate the inner function, \( -x^2 \), to get \(-2x\).
• Multiplying, we find \( f'(x) = e^{-x^2} imes (-2x) = -2xe^{-x^2} \).

The first derivative is crucial as it sets the stage for finding the second derivative and, ultimately, helps identify potential inflection points.

Second Derivative

The second derivative of a function tells us about the concavity of the function, indicating where the function curves upward or downward. To locate inflection points, we need to assess where the second derivative changes sign.

After finding the first derivative \( f'(x) = -2xe^{-x^2} \), we utilize the product rule to find the second derivative. The product rule is essential for differentiating products of two functions.

• The product rule formula is \( (uv)' = u'v + uv' \), where \( u(x) = -2x \) and \( v(x) = e^{-x^2} \).
• Differentiating qualifies \( u'(x) = -2 \) and \( v'(x) = -2xe^{-x^2} \) via the chain rule.
• The second derivative results in \( f''(x) = -2e^{-x^2} + 4x^2e^{-x^2} \).

Furthermore, it simplifies to \( f''(x) = e^{-x^2}(-2 + 4x^2) \), ready to determine where it equals zero, thus identifying inflection points.

Chain Rule

The chain rule is indispensable when differentiating composite functions, functions of functions. For \( f(x) = e^{-x^2} \), understanding and applying the chain rule is essential, especially since it involves the exponential function.

Here’s how the chain rule works in this context:

• Identify the outer function, in this case, \( e^u \).
• The inner function is \( -x^2 \), resulting in the composite function \( u(x) = -x^2 \).
• Derivate the outer function: \( e^u \) becomes \( e^u \) again.
• Differentiate the inner function: \(-x^2\) leads to \(-2x\).
• Multiply these derivatives to obtain: \( e^{-x^2} imes (-2x) = -2xe^{-x^2} \).

Thus, using the chain rule ceaselessly simplifies differentiating and helps find the first derivative accurately.

Product Rule

The product rule helps us when differentiating products of two or more separate functions. In our function's case, \( f'(x) = -2xe^{-x^2} \), we apply this rule to find the second derivative.

Here's how the product rule assists:

• View the derivative as a product: \( u(x) = -2x \) and \( v(x) = e^{-x^2} \).
• The product rule formula: \((uv)' = u'v + uv' \).
• Calculate \( u'(x) = -2 \) and \( v'(x) = -2xe^{-x^2} \) using the chain rule.
• Add these derivatives: \( f''(x) = -2 imes e^{-x^2} + (-2x) imes (-2xe^{-x^2}) \).

This effective application deduces the formula \( f''(x) = e^{-x^2}(-2 + 4x^2) \), highlighting the inflection points through sign changes in the second derivative.