Problem 72 - Single Variable Calculus

Question

a. Show that \(e^{x} \geq 1+x\) if \(x \geq 0\). b. Show that \(e^{x} \geq 1+x+x^{2} / 2\) if \(x \geq 0\). Hint: Show that \(f(x)=e^{x}-1-x-x^{2} / 2\) is increasing for \(x \geq 0\)

Step-by-Step Solution

Answer

In summary, we have shown that \(e^x \geq 1 + x\) and \(e^x \geq 1 + x + \frac{x^2}{2}\) for all \(x \geq 0\) by analyzing the behavior of the functions \(f(x) = e^x - 1 - x\) and \(g(x) = e^x - 1 - x - \frac{x^2}{2}\) and their derivatives, and then evaluating them at the boundary \(x = 0\).

Step 1: Define the function and its derivative

Let's define a function f(x) = e^x - 1 - x. We are trying to show that f(x) ≥ 0 for all x ≥ 0. First, we need to find the derivative of f(x). \(f'(x) = \frac{d}{dx}(e^x - 1 - x)\)

Step 2: Calculate the derivative

To find the derivative, we apply the derivative rules to each term: \(f'(x) = \frac{d}{dx}(e^x) - \frac{d}{dx}(1) - \frac{d}{dx}(x)\) \(f'(x) = e^x - 0 - 1\)

Step 3: Analyze the function's behavior

Since the derivative of f(x) is f'(x) = e^x - 1, it's evident that f'(x) ≥ 0 for all x ≥ 0 as e^x is always greater than or equal to 1 for x greater than or equal to 0. This means the function f(x) is increasing for all x ≥ 0.

Step 4: Evaluate the function at x = 0

To prove the inequality, we will now evaluate f(x) at the boundary x = 0: \(f(0) = e^0 - 1 - 0 = 1 - 1 = 0\) Thus, we have shown that \(e^x \geq 1 + x\) for all x ≥ 0. #b. Proving e^x ≥ 1+x+(x^2)/2 for x ≥ 0#

Step 1: Define the function and its derivative

Now, let's define a new function g(x) = e^x - 1 - x - (x^2)/2. We want to show that g(x) ≥ 0 for all x ≥ 0. First, we need to find the derivative of g(x). \(g'(x) = \frac{d}{dx}(e^x - 1 - x - \frac{x^2}{2})\)

Step 2: Calculate the derivative

To find the derivative, we apply the derivative rules to each term: \(g'(x) = \frac{d}{dx}(e^x) - \frac{d}{dx}(1) - \frac{d}{dx}(x) - \frac{d}{dx}(\frac{x^2}{2})\) \(g'(x) = e^x - 0 - 1 - x\)

Step 3: Analyze the function's behavior

Since the derivative of g(x) is g'(x) = e^x - 1 - x, it's not immediately apparent when g'(x) ≥ 0. However, the hint given suggests that g(x) is increasing for x ≥ 0, which means g'(x) ≥ 0 for x ≥ 0. We can observe that g'(0) = 0 and g'(x) > 0 for x > 0. This confirms that g(x) is indeed an increasing function for all x ≥ 0.

Step 4: Evaluate the function at x = 0

To prove the inequality, we will now evaluate g(x) at the boundary x = 0: \(g(0) = e^0 - 1 - 0 - \frac{0^2}{2} = 1 - 1 = 0\) Thus, we have shown that \(e^x \geq 1 + x + \frac{x^2}{2}\) for all x ≥ 0.

Key Concepts

Exponential Functions

Exponential functions are a fundamental component of many mathematical models used to describe growth and decay. An exponential function is of the form \(a^x\), where \(a>0\) is the base and \(x\) is the exponent. The most important exponential function in calculus is \(e^x\), where \('e'\) is known as Euler's number, approximately equal to 2.718. This function has unique properties that are especially useful in mathematics and the natural sciences.

One key property of the exponential function \(e^x\) is that its derivative with respect to \(x\) is the function itself, that is, \(\frac{d}{dx}e^x = e^x\). This self-replicating feature under differentiation makes \(e^x\) a pivotal concept in solving various problems involving growth rates and changes over time.

Derivatives

In the realm of calculus, the derivative is a fundamental tool that measures how a function's output changes as its input changes. It's typically denoted as \(f'(x)\) or \(\frac{df}{dx}\), representing the rate of change of the function \(f(x)\) with respect to the variable \(x\). To compute derivatives, we apply a series of rules, such as the sum rule, product rule, and chain rule, which allow us to handle more complex expressions.

For example, to differentiate an exponential function like \(e^x\), we use the knowledge that its derivative is simply \(e^x\). For polynomial terms like \(x\) and \(\frac{x^2}{2}\), we apply the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\). In analyzing inequalities involving derivatives, the sign of the derivative gives us insight into whether the function is increasing or decreasing, a concept that heavily ties into function analysis.

Function Analysis

Function analysis involves examining the characteristics of functions to understand their behavior. One of the primary tools in function analysis is investigating the signs of the first and second derivatives. The first derivative, as previously discussed, can tell us whether a function is increasing (if \(f'(x) > 0\)) or decreasing (if \(f'(x) < 0\)). When functions involve exponential terms or polynomials, their increasing or decreasing nature has a significant impact on their graphical representation and the real-world phenomena they model.

The second derivative, although not directly used in this problem, provides information regarding the concavity of a function, influencing how we understand the nature of the function's slope. To prove that a function like \(f(x) = e^x - 1 - x - \frac{x^2}{2}\) is always non-negative, we analyzed its behavior by showing that it is an increasing function from the point \(x = 0\) onward. Function analysis, therefore, allows us to make powerful conclusions about inequalities just with the knowledge of derivatives and the properties of the function concerned.