Problem 15 - Calculus for Biology and Medicine

Question

Show that if \(f(x)\) is a differentiable function for all \(x \in \mathbf{R}\) and with a local minimum at \(x=c\), then \(g(x)=\exp (-f(x))\) has a local maximum at \(x=c\).

Step-by-Step Solution

Answer

\( g(x) = \exp(-f(x)) \) has a local maximum at \( x = c \), as shown by the negative second derivative at this point.

Step 1: Identify Local Minimum Condition

If a function \( f(x) \) has a local minimum at \( x = c \), it means that the first derivative \( f'(c) = 0 \) and the second derivative \( f''(c) > 0 \). The condition \( f'(c) = 0 \) indicates a critical point, and \( f''(c) > 0 \) ensures that the point is indeed a minimum.

Step 2: Express \( g(x) \) and Differentiate

Define \( g(x) = \exp(-f(x)) \). We need to find the critical points of \( g(x) \), so we'll determine \( g'(x) \). Using the chain rule, we have:\[ g'(x) = \frac{d}{dx} \exp(-f(x)) = \exp(-f(x)) \cdot (-f'(x)) = -f'(x) \cdot \exp(-f(x)). \]

Step 3: Evaluate \( g'(x) \) at \( x = c \)

Substitute \( x = c \) into the expression for \( g'(x) \):\[ g'(c) = -f'(c) \cdot \exp(-f(c)). \]Given that \( f'(c) = 0 \), it follows that:\[ g'(c) = 0. \] This shows that \( x = c \) is a critical point for \( g(x) \).

Step 4: Evaluate \( g''(x) \) for Local Extremum

To determine whether this critical point is a maximum, calculate the second derivative \( g''(x) \) using the product rule:\[ g''(x) = \frac{d}{dx}[-f'(x)\exp(-f(x))]. \] First, differentiate \(-f'(x)\) and \(\exp(-f(x))\), resulting in:\[ g''(x) = -f''(x) \cdot \exp(-f(x)) + (f'(x))^2 \cdot \exp(-f(x)). \]

Step 5: Evaluate \( g''(x) \) at \( x = c \)

Substitute \( x = c \) into the expression for \( g''(x) \):\[ g''(c) = -f''(c) \cdot \exp(-f(c)) + (f'(c))^2 \cdot \exp(-f(c)). \]Given \( f'(c) = 0 \) and \( f''(c) > 0 \), it simplifies to:\[ g''(c) = -f''(c) \cdot \exp(-f(c)), \]which is negative because \( f''(c) > 0 \) and \( \exp(-f(c)) > 0 \).

Step 6: Conclusion

Since \( g''(c) < 0 \), the point \( x = c \) is a local maximum for \( g(x) = \exp(-f(x)) \). This negative second derivative confirms concavity downwards at \( x = c \).

Key Concepts

Local minimum

When dealing with differentiable functions like \( f(x) \), understanding local minima is important. A function is said to have a local minimum at a point, \( x = c \), if it is lower than all nearby points on the graph of \( f(x) \). In mathematical terms, this is depicted by two main conditions:

• The first derivative at \( x = c \) is zero, denoted as \( f'(c) = 0 \).
• The second derivative at \( x = c \) is positive, represented as \( f''(c) > 0 \).

These conditions imply that the curve of \( f(x) \) is touching the x-axis at this point without crossing it and is bent upwards, resembling a "smiley face". This characteristic bend suggests that \( x = c \) is a local dip or minimum on the graph.
It's not just about the curve touching the x-axis, but it's about what's happening around it. This is what the second derivative tells us. It shows the curve's "curving behavior," ensuring that \( x = c \) is really a low point rather than just a flat spot.

Chain rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. Suppose we have a function \( g(x) = \exp(-f(x)) \). The chain rule helps us find its derivative, \( g'(x) \), by connecting the derivative of the outer function, \( \exp(u) \), with the derivative of the inner function, \( u = -f(x) \).
Let's break it down:

• Differentiating \( \exp(u) \) with respect to \( u \) gives us \( \exp(u) \).
• Then, differentiating \(-f(x)\) with respect to \( x \) results in \(-f'(x)\).
• Using the chain rule for \( g(x) = \exp(-f(x)) \), we find \( g'(x) = \exp(-f(x)) \cdot (-f'(x)) \).

This ultimately provides \( g'(x) = -f'(x) \cdot \exp(-f(x)) \), effectively linking the rates of change of the inner and outer functions. The chain rule is indispensable for calculating derivatives of nested functions, helping us smoothly transition between related rates of change.

Critical points

Critical points are positions on the graph of a function where the derivative is zero or undefined. For \( f(x) \), having \( f'(c) = 0 \) means \( c \) is a critical point. Critical points are essential because they mark places where the function might change behavior, such as turning from increasing to decreasing or vice versa.
Identifying critical points allows us to determine whether a function's value is at a minimum, maximum, or a saddle point. In the context of \( g(x) = \exp(-f(x)) \):

• Substituting \( f'(c) = 0 \) into \( g'(c) = -f'(c) \cdot \exp(-f(c)) \), we find \( g'(c) = 0 \).
• This confirms that \( x = c \) is a critical point of \( g(x) \).

Knowing when and how the function's graph turns is crucial for analysis. It helps determine potential maxima and minima and guides us toward utilizing tests such as the second derivative test.

Second derivative test

The second derivative test offers insight into whether a critical point of a function is a local minimum, local maximum, or neither. When analyzing a critical point, such as \( x = c \) with a critical point for \( g(x) \), we calculate the second derivative, \( g''(x) \).

• If \( g''(c) > 0 \), \( c \) is a local minimum.
• If \( g''(c) < 0 \), \( c \) is a local maximum.
• If \( g''(c) = 0 \), the test is inconclusive.

For the function \( g(x) = \exp(-f(x)) \), at \( x = c \):

• Using the given that \( f'(c) = 0 \) and \( f''(c) > 0 \), we find \( g''(c) = -f''(c) \cdot \exp(-f(c)) \).
• The result is negative, \( g''(c) < 0 \), which indicates that \( x = c \) is a local maximum for \( g(x) \).

Therefore, knowing the behavior of the second derivative provides conclusive results about the nature of critical points, confirming our earlier analysis of \( g(x) \).