Problem 40 - Calculus

Question

By comparing slopes, prove that if \(x>0\) then (a) \(e^{x}>1+x\) (b) \(e^{-x}>1-x\)

Step-by-Step Solution

Answer

(a) For \(x > 0\), \(e^x\) grows faster than \(1+x\); hence, \(e^x > 1+x\). (b) For \(x > 0\), \(e^{-x}\) decreases slower than \(1-x\); hence, \(e^{-x} > 1-x\).

Step 1: Understand the Functions Involved

We are considering the exponential function \(f(x) = e^x\) and the linear function \(g(x) = 1 + x\) for part (a). For part (b), the relevant functions are \(f(x) = e^{-x}\) and \(g(x) = 1 - x\). Our goal is to compare these using their derivatives, i.e., their slopes.

Step 2: Compute Derivatives for Part (a)

Calculate the derivative of \(f(x) = e^x\), which is \(f'(x) = e^x\). The derivative of \(g(x) = 1 + x\) is \(g'(x) = 1\). For \(x > 0\), \(e^x > 1\). This indicates that for \(x > 0\), the slope of the exponential function is greater than the slope of the linear function.

Step 3: Interpret Derivatives for Part (a)

Since \(f'(x) = e^x > 1 = g'(x)\) for \(x > 0\), the exponential function \(f(x)\) grows faster than the linear function \(g(x)\). This means that starting from \(x = 0\), where \(e^0 = 1\), the function \(f(x) = e^x\) will be steeper and thus will be greater than \(1 + x\) for \(x > 0\). Hence, \(e^x > 1 + x\) when \(x > 0\).

Step 4: Compute Derivatives for Part (b)

Calculate the derivative of \(f(x) = e^{-x}\), which is \(f'(x) = -e^{-x}\). The derivative of \(g(x) = 1 - x\) is \(g'(x) = -1\). For \(x > 0\), \(-e^{-x} > -1\), because \(e^{-x}\) is always positive and decreases as \(x\) increases.

Step 5: Interpret Derivatives for Part (b)

Since \(f'(x) = -e^{-x} > -1 = g'(x)\) for \(x > 0\), the function \(f(x) = e^{-x}\) decreases slower than \(g(x) = 1 - x\). This implies that starting from \(x = 0\), the function \(e^{-x}\) will be greater than \(1 - x\) for \(x > 0\). Therefore, \(e^{-x} > 1 - x\) when \(x > 0\).

Key Concepts

Derivatives

The concept of derivatives is crucial in understanding how functions change. When we talk about a derivative, we refer to the slope of the function at a point. For exponential functions, such as \( f(x) = e^x \) and \( f(x) = e^{-x} \), the derivatives reveal how rapidly these functions increase or decrease.

• The derivative of \( e^x \) is \( e^x \) itself. This means the rate of change of \( e^x \) is proportional to its current value—it grows exponentially.
• The derivative of \( e^{-x} \) is \( -e^{-x} \). Here, the negative sign indicates that the function decreases as \( x \) increases. However, it does so in a manner that becomes less steep rapidly as \( x \) continues to grow.

Comparing these derivatives with those of linear functions \( g(x) = 1+x \) and \( g(x) = 1-x \), whose derivatives are easily seen to be constant at 1 and -1 respectively, helps us understand the stark difference in their behavior. Just knowing these derivatives can tell us a lot about the function's comparative behavior over their domains.

Inequalities

Inequalities allow us to decide which of two quantities is larger or smaller without necessarily finding their exact values. In the context of comparing functions, inequalities can be used to determine how one function compares to another over certain intervals.

• For example, for \( x > 0 \), proving that \( e^x > 1 + x \) involves demonstrating that the exponential function's growth outpaces that of the linear function.
• Similarly, proving \( e^{-x} > 1 - x \) requires showing that the decrease in \( e^{-x} \) is slower than the linear decrease represented by \( 1 - x \).

The use of derivative information is key here. By observing the derivatives of the relevant functions, we can establish how those functions behave relative to each other, thus supporting or demonstrating these inequalities clearly.

Function Comparison

Function comparison is a method used in calculus to evaluate how different functions relate to each other. Here, it is used to compare the exponential functions \( e^x \) and \( e^{-x} \) with their respective linear functions \( 1+x \) and \( 1-x \).

• By observing the derivatives, we see that \( f(x) = e^x \) increases faster than the line \( g(x) = 1 + x \), showing that \( e^x \) will always surpass \( 1 + x \) for any \( x > 0 \).
• Likewise, for \( f(x) = e^{-x} \) and \( g(x) = 1 - x \), the exponential decrease is slower than the linear one. Thus \( e^{-x} \) remains greater than \( 1 - x \) for \( x > 0 \).

This method of comparing functions via their derivatives gives a clear and mathematical way to tackle questions involving inequalities of different functional forms. It provides a deeper analytic insight, which goes beyond mere numerical comparisons.