Problem 23 - Calculus Single Variable

Question

Use the first derivative to determine the intervals on which the given function \(f\) is increasing and on which \(f\) is decreasing. At each point \(c\) with \(f^{\prime}(c)=0,\) use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=2^{x}-x $$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, -\frac{\ln(\ln(2))}{\ln(2)})\) and increasing on \((-\frac{\ln(\ln(2))}{\ln(2)}, \infty)\) with a local minimum at \(x = -\frac{\ln(\ln(2))}{\ln(2)}\).

Step 1: Find the First Derivative

To find out where the function is increasing or decreasing, first compute the derivative of the function. Given: \( f(x) = 2^x - x \). The derivative is \( f'(x) = \frac{d}{dx} (2^x) - \frac{d}{dx} (x) = 2^x \ln(2) - 1 \).

Step 2: Solve for Critical Points

Set the derivative equal to zero to find critical points: \( 2^x \ln(2) - 1 = 0 \). Rearrange to find \( 2^x = \frac{1}{\ln(2)} \). Taking the logarithm of both sides gives \( x \ln(2) = -\ln(\ln(2)) \), so \( x = -\frac{\ln(\ln(2))}{\ln(2)} \).

Step 3: Determine Intervals

Use the critical point to test intervals. Choose test points around the critical number (e.g., to the left and right) to determine the sign of \( f'(x) \). For \( x < -\frac{\ln(\ln(2))}{\ln(2)} \), choose \( x = -1 \), calculate \( f'(-1) = 2^{-1} \ln(2) - 1 < 0 \), implying \( f \) is decreasing. For \( x > -\frac{\ln(\ln(2))}{\ln(2)} \), choose \( x = 0 \), calculate \( f'(0) = 2^0 \ln(2) - 1 > 0 \), implying \( f \) is increasing.

Step 4: Apply the First Derivative Test

The critical point \( c = -\frac{\ln(\ln(2))}{\ln(2)} \) changes from decreasing to increasing, indicating a local minimum at \( f(c) \).

Key Concepts

Increasing Functions

To determine if a function is increasing in a particular interval, one must analyze the signs of its derivative in that interval. The First Derivative Test is vital here. It states that if the derivative, denoted as \( f'(x) \), is greater than zero (i.e., \( f'(x) > 0 \)), the function \( f(x) \) is increasing. This is because a positive slope on a graph indicates an upward trend.

In practice, once you determine critical points by setting the derivative to zero, you take test points within the intervals split by these critical points. If, for example, for an interval \( (a, b) \), the slope test yields \( f'(x) > 0 \) at any test point within \( (a, b) \), then the function is increasing in that whole interval.

For instance, consider our function \( f(x) = 2^x - x \). After finding a critical point at \( c = -\frac{\ln(\ln(2))}{\ln(2)} \), we test \( f'(x) \) at values to the right of \( c \), such as \( x = 0 \). With calculations showing \( f'(0) = 2^0 \ln(2) - 1 > 0 \), it confirms that \( f(x) \) is indeed increasing for \( x > c \). This method ensures that you can reliably detect regions where the function's output rises in response to ascending input values.

Decreasing Functions

Understanding where a function decreases follows a similar path as increasing functions; however, the derivative should be less than zero. Specifically, when \( f'(x) < 0 \), the function is considered decreasing, showing downward movement as you navigate the graph from left to right.

When analyzing intervals for decreasing behavior, the critical points play a crucial role. Choose test points within the intervals around these critical points and assess the sign of \( f'(x) \). If you find \( f'(x) < 0 \) for any selected test point in the interval \( (a, b) \), the function is decreasing throughout this interval.

In our example, apart from calculation around the critical point \( c \), testing \( x = -1 \) gives \( f'(-1) = 2^{-1} \ln(2) - 1 < 0 \). This negative result tells us that \( f(x) = 2^x - x \) is decreasing for all \( x < c \), indicating that as \( x \) becomes smaller, the function value diminishes.

Critical Points

Critical points are the backbone for understanding changes in function behavior between increasing and decreasing intervals. These are the \( x \) values where the derivative equals zero or is undefined, offering points where the function potentially shifts direction.

To find critical points, solve \( f'(x) = 0 \). This gives you key locations to explore further. Critical points act as candidates for local maxima or minima. After identifying these points, you plug them into the First Derivative Test. Analyze the sign changes of \( f'(x) \) around these points to determine if they correspond to peaks or troughs in your graph.

In our problem, the critical point \( c = -\frac{\ln(\ln(2))}{\ln(2)} \) was found by solving \( f'(x) = 0 \). By examining the behavior of the function around \( c \) (decreasing to increasing), we discovered a local minimum at this point. Thus, critical points not only help determine intervals of increase and decrease, but also pinpoint local extrema, shaping our understanding of the function’s overall graph.