Problem 34 - Single Variable Calculus

Question

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ f(x)=e^{x}-x $$

Step-by-Step Solution

Answer

In summary, the function \(f(x)=e^x - x\) has a critical point at \(x=0\), is decreasing on the interval \((-\infty, 0)\), increasing on the interval \((0, \infty)\), and is concave up on \((-\infty, \infty)\). There are no inflection points. By considering these features, we can sketch an approximate graph of the function.

Step 1: Identify the function and its derivatives

First, let's find the first and second derivatives of the function \(f(x)=e^x - x\). First derivative: \[f'(x)=\frac{d}{dx}(e^x - x) = e^x - 1\] Second derivative: \[f''(x)=\frac{d^2}{dx^2}(e^x - 1) = e^x\]

Step 2: Find the critical points

To find the critical points, we need to identify where the first derivative is equal to zero or is undefined. In this case, the first derivative is always defined since it's a continuous function. So, we need to find when it equals zero: \[e^x - 1 = 0\] \[e^x = 1\] This occurs at \(x = 0\). Therefore, there is one critical point: \(x = 0\).

Step 3: Determine where the function is increasing and decreasing

To check whether the function is increasing or decreasing on the intervals determined by the critical points, we will examine the sign of the first derivative. We'll consider the intervals \((-\infty, 0)\) and \((0, \infty)\). On the interval \((-\infty, 0)\): Choose a test point \(x=-1\). Then, \[f'(-1) = e^{-1} - 1 < 0\] So, the function is decreasing on the interval \((-\infty, 0)\). On the interval \((0, \infty)\): Choose a test point \(x=1\). Then, \[f'(1) = e^1 - 1 > 0\] So, the function is increasing on the interval \((0, \infty)\).

Step 4: Find inflection points

To find any inflection points, we will examine the second derivative to determine any changes in concavity. Our second derivative is: \[f''(x) = e^x\] Since the exponential function is always positive and doesn't change the sign, there are no inflection points.

Step 5: Determine the concavity of the function

Since \(f''(x)=e^x > 0\), the function is concave up on its entire domain \((-\infty, \infty)\).

Step 6: Sketch the graph

Now, let's put everything together to sketch the graph of the function \(f(x)=e^x - x\): 1. The function has a critical point at \(x=0\). 2. The function is decreasing on the interval \((-\infty, 0)\) and increasing on the interval \((0, \infty)\). 3. There are no inflection points. 4. The function is concave up on \((-\infty, \infty)\). Based on these features, we can now create a sketch of the function \(f(x)=e^x - x\). Keep in mind that this will not be an exact reproduction of the graph, but rather an approximation to help us visualize the behavior of the function.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus that help us understand how a function is changing at any point. They give us the rate at which the function's value changes as its input changes.

• The **first derivative**, often denoted as \( f'(x) \), tells us if the function is increasing or decreasing at a particular point.
• Here, for the function \( f(x) = e^x - x \), the first derivative is \( f'(x) = e^x - 1 \).
• The **second derivative**, \( f''(x) \), provides insights into the concavity of the function, which is a measure of how the curve bends.
• In this exercise, the second derivative is \( f''(x) = e^x \).

Both derivatives are crucial in curve sketching as they help identify critical points and determine the shape and direction of the graph.

Critical Points

Critical points occur where the first derivative of the function is zero or undefined. These points are significant because they can represent potential local maxima, minima, or points of inflection.

• For the function \( f(x) = e^x - x \), we find the first derivative \( f'(x) = e^x - 1 \).
• Set \( f'(x) = 0 \), which gives us the equation \( e^x = 1 \). Solving this, we find that \( x = 0 \) is a critical point.
• There are no points where \( f'(x) \) is undefined, as the exponential function and \(-x\) are both continuous.

Understanding critical points allows us to divide the number line into intervals where we can analyze the function's behavior in terms of increasing or decreasing trends.

Concavity

Concavity helps us understand the "bend" of a graph.

• If a function's second derivative is positive over an interval, the graph of the function is concave up (like a cup facing upwards).
• If the second derivative is negative, the graph is concave down (like an upside-down cup).
• Here, \( f''(x) = e^x \) is always positive for the function \( f(x) = e^x - x \).
• This means that the graph is concave up on its entire domain \((-\infty, \infty)\).

Understanding concavity helps us predict how the function's slope will change as we move along the curve, which is a crucial step in sketching an accurate graph.

Increasing and Decreasing Intervals

Identifying where a function is increasing or decreasing is essential for sketching its graph. These intervals give information about the function's overall direction.

• A function is **increasing** where its first derivative \( f'(x) \) is positive.
• It is **decreasing** where \( f'(x) \) is negative.
• For this exercise, the first derivative found was \( f'(x) = e^x - 1 \).
• On the interval \((-\infty, 0)\), test with \( x = -1 \), we find \( f'(-1) < 0 \), so the function decreases there.
• On \((0, \infty)\), test with \( x = 1 \), we find \( f'(1) > 0 \), indicating the function increases.

By analyzing these intervals, we can better understand the function's behavior and how it transitions from decreasing to increasing.