Problem 33 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=x+1 / 2^{x} $$

Step-by-Step Solution

Answer

The function has one critical point at \( x = \log_2(\ln(2)) \).

Step 1: Find the First Derivative

To begin analyzing the function, we need to find the first derivative, \( f'(x) \). The given function is \( f(x) = x + \frac{1}{2^x} \). The first derivative of \( f(x) \) is: \[ f'(x) = 1 - \frac{\ln(2)}{2^x} \] where \( \ln(2) \) is the natural logarithm of 2, resulting from differentiating \( \frac{1}{2^x} \).

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. We solve for \( x \) where \( f'(x) = 0 \):\[ 1 - \frac{\ln(2)}{2^x} = 0 \] This simplifies to \( \frac{\ln(2)}{2^x} = 1 \), giving \( 2^x = \ln(2) \), which further simplifies to \( x = \log_2(\ln(2)) \). There are no points where \( f'(x) \) is undefined as the expression is valid for all \( x \).

Key Concepts

First Derivative

The first derivative of a function, denoted as \( f'(x) \), provides valuable insights into the behavior of a function. It indicates the rate at which the function's value is changing at any given point.
To find the first derivative of the function \( f(x) = x + \frac{1}{2^x} \), we use the rule for differentiating the exponential function. The derivative of the linear term \( x \) is straightforward and is 1. For the exponential term, the chain rule gives the derivative as \(-\frac{\ln(2)}{2^x}\).
Thus, the first derivative is given by:
\( f'(x) = 1 - \frac{\ln(2)}{2^x} \).

This derivative tells us whether the function is increasing or decreasing at different \( x \)-values.

Concavity

Concavity describes the curvature of a graph of a function. It indicates whether the graph bends upwards or downwards.

• If a function is concave up, its graph resembles a cup (\(\cup\)), indicating that the slope of the tangent lines is increasing.
• Conversely, if it's concave down, the graph has an arch shape (\(\cap\)), meaning the slope of the tangent lines is decreasing.

To determine concavity, we use the second derivative \( f''(x) \). The sign of \( f''(x) \) tells us the concavity:

• \( f''(x) > 0 \): The function is concave up.
• \( f''(x) < 0 \): The function is concave down.

This analysis helps locate where the graph bends upwards or downwards.

Critical Points

Critical points occur where the first derivative is zero or undefined.
They are important because they can indicate potential maxima, minima, or points of inflection for a function.

• To find these points, we set the first derivative equal to zero and solve.
• For the function \( f(x) = x + \frac{1}{2^x} \), setting \( f'(x) = 0 \) yields:
  \[ 1 - \frac{\ln(2)}{2^x} = 0 \]
• Solving this equation gives \( x = \log_2(\ln(2)) \), which is the critical point.

There are no values for \( x \) that render \( f'(x) \) undefined, so we focus on solving the equation.

Points of Inflection

Points of inflection occur where a function changes its concavity.
At these points, the graph of the function switches from being concave up to concave down, or vice versa.

• The second derivative, \( f''(x) \), helps identify these critical transitions by locating where it changes sign.
• If \( f''(x) \) changes from positive to negative or negative to positive, it indicates a point of inflection.

Points of inflection are crucial as they highlight significant shifts in the shape of a graph, showing where the curve transitions in its bending.