Problem 35 - Calculus Single Variable

Question

In each of Exercises \(35-42,\) calculate the first and second derivatives of the given expression, and classify its local extrema. \(2^{x}-x\)

Step-by-Step Solution

Answer

The local minimum occurs at \(x = \log_2\left(\frac{1}{\ln(2)}\right)\).

Step 1: Find the First Derivative

To find the first derivative of the function \(f(x) = 2^x - x\), apply the following rules:- The derivative of \(2^x\) with respect to \(x\) is \(2^x \ln(2)\).- The derivative of \(-x\) with respect to \(x\) is \(-1\).Thus, the first derivative \(f'(x)\) of the function is:\[ f'(x) = 2^x \ln(2) - 1 \]

Step 2: Find the Second Derivative

Using the derivative found in Step 1, calculate the second derivative:- The derivative of \(2^x \ln(2)\) is \(2^x (\ln(2))^2\) since it remains constant as \(\ln(2)\) is a constant.- The derivative of \(-1\) is \(0\).So the second derivative \(f''(x)\) is:\[ f''(x) = 2^x (\ln(2))^2 \]

Step 3: Find Critical Points

The critical points occur where the first derivative \(f'(x)\) equals zero:\[ 2^x \ln(2) - 1 = 0 \]Solving for \(x\), we get:\[ 2^x \ln(2) = 1 \]\[ 2^x = \frac{1}{\ln(2)} \]\[ x = \log_2\left(\frac{1}{\ln(2)}\right) \]

Step 4: Classify Local Extrema Using the Second Derivative Test

Use the second derivative \(f''(x)\) to determine the nature of the critical point found in Step 3:- Evaluate \(f''(x)\) at the critical point \(x = \log_2\left(\frac{1}{\ln(2)}\right)\):\[ f''(x) = 2^x (\ln(2))^2 > 0 \text{ for all } x \]Because \(f''(x) > 0\), the function is concave upwards, indicating a local minimum at the critical point.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. They provide us the rate of change or the slope of the function at any given point. To find a derivative, we use specific rules of differentiation. For the function we are analyzing, which is given by \( f(x) = 2^x - x \), we calculate the first derivative using the following principles:

• The derivative of an exponential function, such as \( 2^x \), is \( 2^x \ln(2) \). This occurs because the base, \( 2 \), has a natural logarithm \( \ln(2) \).
• The derivative of \(-x\) is simply \(-1\), which follows from standard differentiation rules.

Putting these together, the first derivative of the function, \( f'(x) \), is given by \[ f'(x) = 2^x \ln(2) - 1 \]. This expression tells us about the slopes of the tangent lines for the function at any point \( x \).

Local Extrema

Local extrema refer to the points on a graph where the function reaches its local maximum or minimum values. A local maximum is a peak, while a local minimum is a trough. We find these points by analyzing where the first derivative of the function equals zero.In terms of our function \( f(x) = 2^x - x \), setting the first derivative to zero (\( f'(x) = 2^x \ln(2) - 1 = 0 \)) allows us to find candidate points where local extrema might occur. Solving this equation gives us the possible critical points, which then need classification to determine whether they are minima, maxima, or neither.

Critical Points

Critical points occur where the first derivative, \( f'(x) \), is equal to zero or is undefined. These points are essential in finding where the function might have local maxima or minima.For our function, making the first derivative equal to zero, \( 2^x \ln(2) - 1 = 0 \), we find that:

• First resolve \( 2^x \ln(2) = 1 \).
• Solve for \( x \) to obtain \( x = \log_2\left(\frac{1}{\ln(2)}\right) \).

This value of \( x \) is a critical point. It is where the function changes direction, and we need further investigation, using additional tests, to determine its nature.

Second Derivative Test

The second derivative test gives us information about the concavity of the function at critical points, which helps determine whether they are maxima, minima, or points of inflection.For a function \( f(x) \), the second derivative is \( f''(x) \). Evaluating this derivative at a critical point can tell us:

• If \( f''(x) > 0 \), the function is concave upwards, and the critical point is a local minimum.
• If \( f''(x) < 0 \), the function is concave downwards, indicating a local maximum.

In our exercise, the second derivative is calculated as \( f''(x) = 2^x (\ln(2))^2 \), which is always positive, showing that the function is concave upward at all critical points, confirming a local minimum. Thus, at \( x = \log_2\left(\frac{1}{\ln(2)}\right) \), the function \( 2^x - x \) indeed hits a local minimum.