Question

\(e^{\pi}>\pi^{\epsilon}\) Prove that \(e^{\pi}>\pi^{e}\) by first finding the maximum value of \(f(x)=\ln x / x\)

Short answer

Question: Prove that \(e^{\pi} > \pi^{e}\) by finding the maximum value of \(f(x) = \ln x / x\). Solution: By finding the maximum of the function \(f(x) = \ln x / x\), which occurs at \(x \approx 2.72\), and showing that \(f(\pi) < f(e)\), we determined that \(e^{\pi} > \pi^{e}\).

Step-by-step solution

Understand the given function

We are given the function \(f(x)=\ln x / x\). We have to find the maximum value of this function. To do this, we'll use the first and second derivative tests to find the critical points and determine their nature (i.e., whether they are maximum points, minimum points, or points of inflection).

Find the first derivative of the function

First, find the derivative of the function, \(f'(x)\), with respect to \(x\): \(f'(x) = \frac{d}{dx}\left(\frac{\ln x}{x}\right)\) To find the derivative, we can apply the Quotient Rule, which states: \(\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\) where \(u=\ln x\) and \(v=x\). Let's find \(u',v',uv'\): \(u'=\frac{d}{dx}(\ln x) = \frac{1}{x}\) \(v'= \frac{d}{dx}(x) = 1\) \(uv' = (\ln x)(1)= \ln x\) Now, we can apply the Quotient Rule: \(f'(x) = \frac{u'v - uv'}{v^2} = \frac{(\frac{1}{x})(x) - (\ln x)(1)}{x^2}= \frac{x - \ln x}{x^2}\)

Find the critical points of the function

Next, find the critical points of the function, \(f(x)\) by setting \(f'(x) = 0\). These critical points will help us determine the maximum values of \(f(x)\): \(f'(x) = \frac{x - \ln x}{x^2} = 0\) Since the denominator cannot equal zero, we focus on the numerator: \(x - \ln x = 0\) This equation cannot be solved algebraically, but we can use numerical methods to find its solution. In this case, the critical point \(x \approx 2.72\).

Find the second derivative of the function

Now, find the second derivative of the function, \(f''(x)\), which will help us determine if the critical point is a maximum or minimum: \(f''(x) = \frac{d}{dx}\left(\frac{x - \ln x}{x^2}\right)\) To do this, we'll apply the Quotient Rule again: \(f''(x) = \frac{(1 - \frac{1}{x})x^2 - 2x(x - \ln x)}{x^4}= \frac{x^2 - 2x^2 + 2x\ln x}{x^4} = \frac{-x^2 + 2x\ln x}{x^4}\)

Determine the nature of the critical point

Using the second derivative test, we'll plug the critical point of \(x \approx 2.72\) into \(f''(x)\): \(f''(2.72) \approx \frac{-2.72^2 + 2(2.72)(\ln 2.72)}{2.72^4} < 0\) Since \(f''(x)\) is negative at \(x\approx2.72\), the critical point is a maximum. Now, we can say that the maximum value of \(f(x)\) occurs at \(x\approx 2.72\).

Prove that \(e^{\pi} > \pi^{e}\)

With the maximum value of \(f(x)\) at \(x \approx 2.72\), note that \(\pi>e\). Since the logarithmic function is increasing for \(x>0\), which means that for \(x>y>0\), we have \(\ln x > \ln y\). Therefore, for our case: \(\ln\pi > \ln e\) Now, we have: \(f(\pi) = \frac{\ln\pi}{\pi} < f(e) = \frac{\ln e}{e}\) By multiplying both sides by \(e^{\pi}>0\) and by noting that \(\frac{\ln e}{e} = 1\): \(\pi^{e} < e^{\pi}\) Consequently, \(e^{\pi} > \pi^{e}\) We have now shown that \(e^{\pi} > \pi^{e}\) by first finding the maximum value of \(f(x)=\ln x / x\).

Key concepts

Derivative Test

The Derivative Test is a fundamental concept in calculus used to determine the nature of critical points on a graph. By examining the first and second derivatives of a function, we gain insight into the behavior of the function at critical points.

• First Derivative Test: This test involves finding the first derivative of a function and setting it to zero to identify critical points. Once we locate these points, we check the sign of the derivative on both sides. A change from positive to negative indicates a local maximum, while a change from negative to positive suggests a local minimum.
• Second Derivative Test: After finding a critical point using the first derivative, we can deploy the second derivative to determine whether the point is a local maximum, minimum, or a point of inflection. If the second derivative is positive at the critical point, it's a local minimum; if negative, it's a local maximum. If it equals zero, the test is inconclusive.

These derivative tests were crucial in our original problem to confirm that the critical point we found was indeed a maximum, which helped prove that one value was greater than the other.

Critical Points

Critical points are specific values on a function's graph where the first derivative equals zero or is undefined. They represent potential maxima, minima, or points of inflection. Identifying these points is essential in understanding the overall shape and behavior of a function.
For example, consider the task of finding the maximum value of the function \(f(x)=\frac{\ln x}{x}\). The critical point emerges at \( x \approx 2.72 \) as determined by solving the equation \( x - \ln x = 0 \) numerically. While this equation doesn't have a straightforward solution, numerical methods provide a near approximation.
Once we find the critical points, we employ the second derivative test to establish their nature, confirming if they are points of maxima or minima. This process allows us to ascertain that the maximum occurs at the critical point, playing a crucial role in the problem.

Logarithmic Function

Logarithmic functions, like the natural logarithm \(\ln x\), are essential in many areas of mathematics and science. They are the inverse functions of exponentials and are commonly used to solve equations involving exponential growth or decay.

• The natural logarithm \(\ln x\) is defined as the power to which the base \(e\) must be raised to produce a given number \(x\).
• Properties of logarithms help simplify complex expressions and are key in calculus for differentiation and integration tasks.

In the original problem, the logarithmic function \(\ln x\) came into play as part of the expression \(f(x) = \frac{\ln x}{x}\). The behavior of \(\ln x\) is essential to finding and understanding the maximum value of \(f(x)\), ultimately aiding in demonstrating the inequality \(e^\pi > \pi^e\).

Quotient Rule

The Quotient Rule is a technique for differentiating functions expressed as a ratio of two functions, \(\frac{u}{v}\). It is particularly useful when dealing with complex rational expressions in calculus.
The rule is stated as follows: If \(y = \frac{u}{v}\), then the derivative \(y'\) with respect to \(x\) is \(\frac{u'v - uv'}{v^2}\), where \(u\) and \(v\) are differentiable functions of \(x\).

• This formula helps us find the derivative of any quotient of two differentiable functions.
• It plays a crucial role in problems where functions are defined in terms of ratios, much like \(f(x) = \frac{\ln x}{x}\) in our exercise.

In the given solution, the Quotient Rule was applied to derive \(f'(x)\), which helped identify the critical points essential for applying the Derivative Test. Understanding and applying this rule accurately ensures precise computation of derivatives in such contexts.