Problem 34 - Applied Calculus

Question

Let \(g(x)=x-k e^{x}\), where \(k\) is any constant. For what value(s) of \(k\) does the function \(g\) have a critical point?

Step-by-Step Solution

Answer

For a critical point, \( k = 0 \).

Step 1: Understanding Critical Points

A critical point of the function occurs where the derivative is zero or undefined. We need to find the derivative of the function and set it to zero to locate critical points.

Step 2: Differentiating the Function

The given function is \( g(x) = x - k e^{x} \). To find the derivative, apply the basic differentiation rules. The derivative is: \( g'(x) = 1 - k e^{x} \).

Step 3: Setting the Derivative to Zero

To find the critical points, set the derivative equal to zero: \( 1 - k e^{x} = 0 \). Solving for \( k \) will give \( k = \frac{1}{e^{x}} \).

Step 4: Analyzing the Equation

For \( g(x) \) to have at least one critical point, \( \frac{1}{e^{x}} \) must be constant for some \( x \). However, this requires specific values of \( x \) for \( k \) to be a well-defined constant. A critical point is guaranteed when \( k = 0 \) because any \( k > 0 \) would require \( x \) to vary continuously.

Key Concepts

Differentiation Rules

Understanding differentiation rules is crucial in finding critical points for functions, especially when dealing with complex expressions. Differentiation allows us to compute the derivative, which provides the rate of change of a function. In the problem with the function \( g(x) = x - k e^{x} \), applying differentiation helps us identify where critical points could potentially occur.

• The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
• For constant coefficients, the derivative of \( cx \) is simply \( c \).
• The derivative of the exponential function \( e^{x} \) remains \( e^{x} \).

When differentiating \( g(x) \), we note that the derivative of \( x \) is \( 1 \), and the derivative of \( -k e^{x} \) is \( -k e^{x} \). Combining these results, we get the derivative \( g'(x) = 1 - k e^{x} \).
This outcome is pivotal as setting this derivative to zero helps locate potential critical points, where the function's slope is zero.

Exponential Functions

Exponential functions are functions of the form \( f(x) = a e^{bx} \), where \( e \) is the base of natural logarithms, approximately equal to 2.718. Such functions exhibit unique properties crucial for solving calculus problems.
In the given function \( g(x) = x - k e^{x} \), exponential functions play a vital role due to the presence of \( e^{x} \).

• Exponential functions grow rapidly as \( x \) increases.
• They have a special property where the derivative of \( e^{x} \) remains \( e^{x} \).

In our context, understanding how \( e^{x} \) behaves is necessary. It helps in setting up equations to find critical points, where the function \( g(x) \) may change direction. Specifically, \( e^{x} \) does not allow for zero as it never equals zero naturally. This aspect is why you see it affecting our solution for \( k = \frac{1}{e^{x}} \).
The exponential component influences how we reason about the derivation and necessary computations to find critical points.

Solving Equations

In mathematics, solving equations means finding the values of variables that satisfy the equation. For the given function \( g(x) = x - k e^{x} \), solving the equation involves searching for possible values of \( k \) that allow for an existing critical point.

• Start by setting the derivative equal to zero: \( 1 - k e^{x} = 0 \).
• This leads to rearranging the equation to solve for \( k \): \( k = \frac{1}{e^{x}} \).

For \( g(x) \) to have a critical point, \( \frac{1}{e^{x}} \) must be constant for some \( x \), which implies \( k \) must remain consistent over the interval or set in question.
However, \( k = 0 \) provides a guaranteed critical point, as it simplifies the function to \( g(x) = x \), which is linear and has a constant slope of zero. Thus, solving such equations and understanding possible values for \( k \) aids in determining where critical points might occur effectively.