Problem 56 - Calculus Single Variable

Question

Find the minimum value \(m\) of \(f(x)=x-\ln (x)\). Use the inequality \(x-\ln (x) \geq m\) to deduce that \(\ln (1+x) \leq x\) for \(-1

Step-by-Step Solution

Answer

The minimum of \(f(x)\) is 1, occurring at \(x=1\); no real \(x\) solves \(\ln(x)=x\).

Step 1: Differentiate the function

Calculate the derivative of the function given by \(f(x) = x - \ln(x)\). The derivative is found using the power rule and the derivative of a logarithm:\[f'(x) = 1 - \frac{1}{x}\]

Step 2: Find critical points

Set the derivative equal to zero to find critical points:\[1 - \frac{1}{x} = 0\]Solving this equation gives:\[x = 1\]

Step 3: Analyze critical points

Evaluate \(f(x)\) at the critical point \(x = 1\) and at any endpoints if applicable. Since \(x = 1\) is within the domain of \(x > 0\):\[f(1) = 1 - \ln(1) = 1\]

Step 4: Use second derivative test

Compute the second derivative \(f''(x)\):\[f''(x) = \frac{1}{x^2}\]Since \(f''(x) > 0\) for \(x > 0\), \(f(x)\) is concave up, confirming \(x = 1\) is a minimum point.

Step 5: Relate to inequality

From \(f(x) \geq 1\), we get:\[x - \ln(x) \geq 1\]Rearranging gives:\[-\ln(x) \geq 1 - x \quad \text{or} \quad \ln(x) \leq x - 1\]Letting \(x = 1 + y\) with \(-1 < y\):\[\ln(1+y) \leq y\] which is exactly what was needed.

Step 6: Check for \(\ln(x) = x\)

Assume \(\ln(x) = x\) and see if any solutions exist. Rearrange the equation to compare with the earlier conclusion:\[x - \ln(x) = 0\]Since \(x - \ln(x) \geq 1\) for \(x > 0\), the only solution must fall outside the valid domain. Therefore, there is no real \(x\) such that \(\ln(x) = x\).

Key Concepts

Derivative Test

The derivative test is a fundamental tool in calculus used to identify the local extrema of a function. To apply this test, the first derivative of the function is calculated, which reflects the slope of the tangent line to the graph at any point. For the function given, \( f(x) = x - \ln(x) \), the derivative is determined as:\[ f'(x) = 1 - \frac{1}{x} \]

• If the derivative \( f'(x) = 0 \), the point is a potential location for a local maximum or minimum.
• If \( f'(x) > 0 \) in an interval, the function is increasing in that region.
• If \( f'(x) < 0 \) in an interval, the function is decreasing in that region.

By solving \( 1 - \frac{1}{x} = 0 \), we found the critical point at \( x = 1 \). This point is significant because it gives us a specific location to further analyze for identifying minimum or maximum values.

Function Critical Points

Critical points of a function are where the function's derivative is either zero or undefined, revealing crucial turning points. In our exercise, after computing the derivative \( f'(x) = 1 - \frac{1}{x} \), we set it to zero to find:\[ 1 - \frac{1}{x} = 0 \Rightarrow x = 1 \]This critical point suggests a possible minimum or maximum at \( x = 1 \). Evaluating the original function \( f(x) \) at this point confirms:\[ f(1) = 1 - \ln(1) = 1 \]This shows that when \( x = 1 \), the function achieves a value of 1. Critical points, like \( x = 1 \), help us understand where the function's behavior changes, such as switching from increasing to decreasing or vice versa.

Inequality Proving

Inequality proving involves showing that one expression is consistently greater than or less than another. In this exercise, we employ the result \( x - \ln(x) \geq 1 \), derived from examining our function's behavior:\[ \ln(x) \leq x - 1 \]This is rearranged to demonstrate:\[ \ln(1 + y) \leq y \text{ for } -1 < y \text{ by substituting } x = 1 + y \]

• This inequality holds because \( x - \ln(x) \) is minimized at \( x = 1 \) and cannot go below 1.
• The inequality serves as a fundamental relationship in analyzing calculations involving natural logarithms.

Such inequalities are crucial, not only for their immediate results but for understanding the behavior and limits of functions across different regions.

Concavity and Convexity

Concavity and convexity describe the curvature of a graph and are examined through the second derivative test. In this exercise, after finding the second derivative:\[ f''(x) = \frac{1}{x^2} \]

• If \( f''(x) > 0 \), the graph is concave up, suggesting the function is shaped like a bowl, indicating a local minimum.
• If \( f''(x) < 0 \), the graph is concave down, suggesting a local maximum.

For \( x > 0 \), since \( f''(x) = \frac{1}{x^2} \) is always positive, our function \( f(x) \) is concave up in its entire domain of \( x > 0 \). This reassures us that the critical point found at \( x = 1 \) is indeed a minimum. Understanding concavity and convexity is vital for predicting how the graph behaves, guiding interpretations of function trends and extremities.