Problem 45 - Calculus Early Transcendentals

Question

(a) Show that \(\ln (x+1) \leq x\) if \(x \geq 0\) (b) Show that \(\ln (x+1) \geq x-\frac{1}{2} x^{2}\) if \(x \geq 0\). (c) Confirm the inequalities in parts (a) and (b) with a graphing utility.

Step-by-Step Solution

Answer

For (a) \( \ln(x+1) \leq x \) and (b) \( \ln(x+1) \geq x - \frac{1}{2}x^2 \), verified analytically and graphically.

Step 1: Analyze the function for part (a)

Consider the function \( f(x) = x - \ln(x+1) \). We want to prove that \( f(x) \geq 0 \) for \( x \geq 0 \). Begin by taking the derivative: \( f'(x) = 1 - \frac{1}{x+1} \).

Step 2: Derivative Analysis for part (a)

Calculate the derivative: \( f'(x) = 1 - \frac{1}{x+1} = \frac{x}{x+1} \geq 0 \) for \( x \geq 0 \). This means the function \( f(x) \) is non-decreasing on \( x \geq 0 \).

Step 3: Initial Value for part (a)

Evaluate \( f(0) = 0 - \ln(1) = 0 \). Since \( f(x) \) is non-decreasing and \( f(0) = 0 \), \( f(x) \geq 0 \) for \( x \geq 0 \). Thus, \( \ln(x+1) \leq x \).

Step 4: Analyze the function for part (b)

Consider the function \( g(x) = \ln(x+1) - \left(x - \frac{1}{2}x^2\right) \). We aim to show \( g(x) \geq 0 \) for \( x \geq 0 \). Evaluate the derivative: \( g'(x) = \frac{1}{x+1} - 1 + x \).

Step 5: Simplify the derivative for part (b)

Simplify \( g'(x): \frac{1}{x+1} - 1 + x = \frac{1 - (x+1) + x(x+1)}{x+1} = \frac{x^2}{x+1} \geq 0 \) for \( x \geq 0 \). Therefore, \( g(x) \) is non-decreasing.

Step 6: Initial Value for part (b)

Evaluate \( g(0) = \ln(1) - (0 - \frac{1}{2} \cdot 0^2) = 0 \). Since \( g(x) \) is non-decreasing and \( g(0) = 0 \), \( g(x) \geq 0 \) and hence \( \ln(x+1) \geq x - \frac{1}{2}x^2 \).

Step 7: Graphing Utility Confirmation

Plot the functions \( y_1 = \ln(x+1) \), \( y_2 = x \), and \( y_3 = x - \frac{1}{2}x^2 \) on a graphing calculator or software. Confirm graphically that \( \ln(x+1) \) lies below the line \( y = x \) for part (a) and above the parabola \( y = x - \frac{1}{2}x^2 \) for part (b) for \( x \geq 0 \).

Step 8: Conclusion

By showing both analytically and graphically, we confirm that \( \ln(x+1) \leq x \) and \( \ln(x+1) \geq x - \frac{1}{2}x^2 \) for \( x \geq 0 \).

Key Concepts

Derivative Analysis

To understand the inequalities involving the natural logarithm function, we can start by analyzing the derivatives of the two functions given in the exercise.

• For part (a), we consider the function \( f(x) = x - \ln(x+1) \). Our goal is to show that \( f(x) \geq 0 \) for \( x \geq 0 \). This means \( \ln(x+1) \leq x \).
• To do this, we calculate the derivative: \( f'(x) = 1 - \frac{1}{x+1} \).
• Simplifying, we have \( \frac{x}{x+1} \geq 0 \) for \( x \geq 0 \), indicating \( f(x) \) is non-decreasing. With \( f(0) = 0 \), it confirms that \( f(x) \geq 0 \) for \( x \geq 0 \).

Similarly, for part (b), we look at the function \( g(x) = \ln(x+1) - \left(x - \frac{1}{2}x^2\right) \).

• The derivative here is \( g'(x) = \frac{1}{x+1} - 1 + x \).
• Simplifying gives \( \frac{x^2}{x+1} \geq 0 \) for \( x \geq 0 \), also showing the function is non-decreasing.
• With \( g(0) = 0 \), we ensure \( g(x) \geq 0 \) for \( x \geq 0 \), proving \( \ln(x+1) \geq x - \frac{1}{2}x^2 \).

Graphical Confirmation

Graphing the functions is another way to visually assert the inequalities. By using a graphing utility such as a calculator or software tool, you can plot the functions to see their relationships clearly.

• For part (a), plot \( y_1 = \ln(x+1) \) and \( y_2 = x \). What you expect to see is that the graph of \( y = \ln(x+1) \) lies below the line \( y = x \) for \( x \geq 0 \).
• For part (b), plot \( y_1 = \ln(x+1) \) and \( y_3 = x - \frac{1}{2}x^2 \). Here, the graph of \( y = \ln(x+1) \) should lie above the curve \( y = x - \frac{1}{2}x^2 \) for \( x \geq 0 \).
• This visual confirmation through graphs helps consolidate the analytical results and provides a clear depiction of why the inequalities hold true.

Logarithmic Functions

The natural logarithmic function \( \ln(x+1) \) is central to these inequalities. Understanding this function allows you to grasp why certain inequalities hold.

• The function \( \ln(x+1) \) is defined for \( x \geq -1 \) and is continuous and increasing in this interval.
• Its behavior near \( x = 0 \) is particularly important because the properties of the function derive from its derivative \( \frac{1}{x+1} \). This derivative shows how rapidly \( \ln(x+1) \) increases.
• For the function \( \ln(x+1) \), at \( x = 0 \), the function returns \( \ln(1) = 0 \), which serves as the initial condition for the inequalities.
• This foundational knowledge helps in analyzing the increasing or decreasing nature of related functions, thus supporting the inequalities with the logarithmic expressions.