Problem 24 - 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. $$ f(x)=2^{x}-4^{x} $$

Step-by-Step Solution

Answer

Function \( f(x) \) is increasing on \((-\infty, -1)\) and decreasing on \((-1, \infty)\), with a local maximum at \( x = -1 \).

Step 1: Find the First Derivative

To find the derivative of the function \( f(x) = 2^x - 4^x \), use the exponential derivative rule: \( \frac{d}{dx} a^x = a^x \ln(a) \). Therefore, the derivative is \( f'(x) = 2^x \ln(2) - 4^x \ln(4) \).

Step 2: Set the First Derivative to Zero

Find the critical points by solving \( f'(x) = 0 \), i.e., \( 2^x \ln(2) - 4^x \ln(4) = 0 \). Rewriting \( 4^x \) as \( (2^2)^x = (2^x)^2 \), the equation becomes \( 2^x \ln(2) = (2^x)^2 \ln(4) \). Factor out \( 2^x \), giving \( 2^x(\ln(2) - 2^x \ln(4)) = 0 \). Therefore, \( 2^x = 0 \) is impossible, so solve \( \ln(2) = 2^x \ln(4) \), leading to \( 2^x = \frac{\ln(2)}{\ln(4)} \).

Step 3: Simplify and Solve for Critical Points

Since \( \ln(4) = 2\ln(2) \), we have \( 2^x = \frac{\ln(2)}{2\ln(2)} = \frac{1}{2} \). Taking the log base 2, \( x = \log_2\left(\frac{1}{2}\right) = -1 \). Thus, the critical point \( x = -1 \).

Step 4: First Derivative Test for Increasing and Decreasing Intervals

To apply the First Derivative Test, check the signs of \( f'(x) \) around the critical point \( x = -1 \). If \( x < -1 \), pick \( x = -2 \): \( f'(-2) = 2^{-2}\ln(2) - 4^{-2}\ln(4) > 0 \), implying \( f'(x) > 0 \). Hence, \( f(x) \) is increasing on \((-\infty, -1)\).For \( x > -1 \), pick \( x = 0 \): \( f'(0) = 2^0\ln(2) - 4^0\ln(4) = \ln(2) - \ln(4) < 0 \), implying \( f'(x) < 0 \). Hence, \( f(x) \) is decreasing on \((-1, \infty)\).

Step 5: Determine Local Maximum or Minimum

Since \( f'(x) \) changes from positive to negative at \( x = -1 \), it indicates a local maximum at this point according to the First Derivative Test.

Key Concepts

Critical Points

In mathematics, critical points of a function are values where the derivative is zero or undefined. These points are essential because they offer potential locations for local maxima, minima, or points of inflection. For the function given, \[ f(x) = 2^x - 4^x, \]we begin by finding the first derivative: \[ f'(x) = 2^x \ln(2) - 4^x \ln(4). \]Setting this derivative to zero helps us identify critical points. A critical point is found when \[ 2^x \ln(2) - 4^x \ln(4) = 0. \]Through some algebraic manipulation, as shown in the solution, we find that the only critical point is at \[ x = -1. \]This critical point helps us determine how the function behaves around this value, leading us to further analysis of increasing and decreasing intervals and local maxima or minima.

Increasing and Decreasing Intervals

Identifying intervals where a function is increasing or decreasing is crucial in understanding its overall behavior. Once we've located the critical point from the first derivative, we can deduce the nature of the function around this point. For the function \[ f(x) = 2^x - 4^x, \]analyze the sign of the derivative, \[ f'(x) = 2^x \ln(2) - 4^x \ln(4), \]around the critical point \[ x = -1. \]Evaluate the derivative at points less than and greater than \[ x = -1. \]

• For \( x < -1 \), such as \( x = -2 \), the derivative is positive, indicating that \( f(x) \) is increasing on the interval \((-\infty, -1)\).
• For \( x > -1 \), such as \( x = 0 \), the derivative is negative, showing that \( f(x) \) is decreasing on the interval \((-1, \infty)\).

Utilizing these intervals, we can visualize the slope of the function and predict its behavior.

First Derivative Test

The First Derivative Test is a useful tool to determine whether a critical point is a local maximum, a local minimum, or neither. By observing the change in the sign of the first derivative around the critical point, valuable insights into the function can be discovered.
For \( f(x) = 2^x - 4^x, \) the critical point was found to be at \( x = -1. \) By checking the sign of \( f'(x) = 2^x \ln(2) - 4^x \ln(4), \)around \( x = -1, \)we make the following observations:

• As \( x \) approaches \(-1 \) from the left, \( f'(x) \) is positive, indicating increasing behavior.
• As \( x \) crosses \(-1 \) and moves to the right, \( f'(x) \) becomes negative, showing decreasing behavior.

When the derivative changes from positive to negative, the function has a local maximum at \( x = -1. \) This tells us that at \( x = -1, \)\( f(x) \) peaks before beginning to decline. Understanding these changes helps in visualizing the function's graph and interpreting the nature of the critical points correctly.