Problem 22 - Calculus for Biology and Medicine

Question

In Problems , find c such that \(f^{\prime}(c)=0\) and determine whether \(f(x)\) has a local extremum at \(x=c .\) $$ f(x)=e^{-x^{2}} $$

Step-by-Step Solution

Answer

At \( x = 0 \), \( f(x) \) has a local maximum because \( f''(0) < 0 \).

Step 1: Find the Derivative

To find where the derivative is zero, we first need to find the derivative of the function. Given the function is \( f(x) = e^{-x^2} \), we use differentiation techniques. The derivative \( f'(x) \) is found using the chain rule, where \( u(x) = -x^2 \) and \( v(u) = e^u \). We get: \[ f'(x) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2} \].

Step 2: Set the Derivative to Zero

Now that we have the derivative \( f'(x) = -2xe^{-x^2} \), we set it equal to zero to find the critical points: \[ -2xe^{-x^2} = 0 \]. Since \( e^{-x^2} eq 0 \) for all real \( x \), the equation simplifies to \( -2x = 0 \).

Step 3: Solve for Critical Points

Solving \( -2x = 0 \) gives us \( x = 0 \). Thus, the critical point is at \( x = 0 \).

Step 4: Determine Local Extremum

To determine if there is a local extremum at \( x = 0 \), we analyze the second derivative \( f''(x) \). First, compute \( f''(x) \) from \( f'(x) = -2xe^{-x^2} \) using the product and chain rule:\[ f''(x) = -2e^{-x^2} + 4x^2e^{-x^2} = 2e^{-x^2}(2x^2 - 1) \]. Evaluate \( f''(0) \):\[ f''(0) = 2e^0(-1) = -2 \]. Since \( f''(0) < 0 \), \( f \) has a local maximum at \( x = 0 \).

Key Concepts

Derivative

The derivative of a function is a fundamental concept in calculus that represents the rate at which a function is changing at any given point. Think of it as a mathematical tool that tells you how steep a function's graph is at any particular spot. For the function \( f(x) = e^{-x^2} \), finding the derivative involves applying the chain rule since the function is a composite of two functions: the exponential function and a polynomial. This ensures we correctly capture how the function behaves through its complex structure. Differentiating, we find the derivative:

• First, recognize the inner function \( u(x) = -x^2 \) and the outer function \( v(u) = e^u \).


• Apply the chain rule: the derivative of the outer function times the derivative of the inner function.

Therefore, \( f'(x) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2} \). This derivative helps us find critical points, where the function stops increasing or decreasing momentarily.

Local Extrema

Local extrema are the peaks and troughs found in a graph, representing high or low points in a particular neighborhood of the function. To find local extrema, you first calculate where the derivative is zero or undefined, signaling a potential change in direction. With the derivative \( f'(x) = -2xe^{-x^2} \), we set it to zero

• Solve \( -2x = 0 \), leading to the critical point \( x = 0 \).

This tells us that the function potentially has a local extremum at this point. However, we need further testing to confirm if it is a maximum, minimum, or neither.

Chain Rule

The chain rule in calculus is a technique used to differentiate composite functions. When you have a function nested within another function, the chain rule helps break down the differentiation process. For \( f(x) = e^{-x^2} \), the chain rule is essential because:

• The function \( e^{-x^2} \) involves an exponential function where the exponent is itself a function \( -x^2 \).


• Using the chain rule, differentiate the outer function \( e^u \) and multiply it by the derivative of \( u = -x^2 \), which is \( -2x \).

This results in the derivative \( f'(x) = -2xe^{-x^2} \). The chain rule allows for precise differentiation of complex expressions by systematically tackling one function at a time.

Second Derivative Test

The second derivative test is a method used to classify critical points found from the first derivative test into local maxima or minima. After finding a critical point using the first derivative, such as \( x = 0 \) from \( f'(x) = 0 \), the second derivative \( f''(x) \) helps determine the nature of this point:

• If \( f''(x) > 0 \), it indicates a local minimum.


• If \( f''(x) < 0 \), it means a local maximum.


• If \( f''(x) = 0 \), the test is inconclusive.

For \( f(x) = e^{-x^2} \), the second derivative is \( f''(x) = 2e^{-x^2}(2x^2 - 1) \). Evaluating at \( x = 0 \), we find \( f''(0) = -2 \), which is less than zero, confirming a local maximum at \( x = 0 \). This test effectively categorizes the critical point's significance, ensuring clarity in the function's behavior.