Problem 21 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Use the function \(f(x)=x^{\frac{1}{x}}, x>0\) to determine the bigger of the two numbers \(e^{\pi} \& \pi^{e}\).

Step-by-Step Solution

Answer

The bigger number between \(e^{\pi}\) and \(\pi^{e}\) is \(e^{\pi}\), as determined by analyzing the behavior of the function \(f(x) = x^{\frac{1}{x}}\) around its critical point.

Step 1: Find the critical points of the function

To find the critical points, we need to take the derivative of the function and set it to zero: \[f(x) = x^{\frac{1}{x}}\] To do this, let's first apply the natural logarithm to both sides: \[\ln{f(x)} = \ln{x^{\frac{1}{x}}}\] Using properties of logarithms, we get: \[\ln{f(x)} = \frac{1}{x}\ln{x}\] Now let's take the derivative using chain rule and implicit differentiation: \[\frac{f'(x)}{f(x)} = \frac{-\ln{x} + 1}{x^2}\] To find the critical points, set the derivative equal to zero: \[f'(x) = 0 \Rightarrow \frac{-\ln{x} + 1}{x^2} = 0\] We see that \(-\ln{x} + 1 = 0 \Rightarrow \ln{x} = 1\) This implies that \(x = e\). So, the only critical point is \(x=e\).

Step 2: Determine the behavior of the function around the critical point

Now that we've found the critical point, we want to determine how the function behaves around this point. We can do this by looking at the sign of the derivative on either side of \(x=e\). If \(f'(x) > 0\), then the function is increasing. If \(f'(x) < 0\), then the function is decreasing. We don't need to evaluate the whole expression for the derivative, we can consider only the numerator part (-ln(x) + 1). For \(x < e\): \(-\ln{x} + 1 > 0\) For \(x > e\): \(-\ln{x} + 1 < 0\) This means that the function is increasing for \(x < e\) and decreasing for \(x > e\).

Step 3: Compare the values of the function at e and π

Now we can compare the values of the function at \(x = e\) and \(x = \pi\): Remember that \(f(x) = x^{\frac{1}{x}}\) At \(x = e\): \[f(e) = e^{\frac{1}{e}}\] At \(x = \pi\): \[f(\pi) = \pi^{\frac{1}{\pi}}\] Since the function is increasing for \(x < e\) and decreasing for \(x > e\), we can say that: \[f(e) > f(\pi)\] Which implies: \[e^{\frac{1}{e}} > \pi^{\frac{1}{\pi}}\] Now, let's raise both sides of the inequality to the \(e\pi\) power: \[\left(e^{\frac{1}{e}}\right)^{e\pi} > \left(\pi^{\frac{1}{\pi}}\right)^{e\pi}\] This simplifies to: \[e^{\pi} > \pi^{e}\] So, \(e^{\pi} > \pi^{e}\).

Key Concepts

Critical Points

In calculus, critical points are the values of \(x\) at which a function's derivative is zero or undefined. These points help us understand where functions might have a maximum, minimum, or a change in the direction of their slopes.

To find a critical point, follow these steps:

• First, calculate the derivative of the function.
• Next, set this derivative equal to zero and solve for \(x\).
• Check if any solutions also make the derivative undefined, as these can also be critical points.

In the given exercise, the function \(f(x) = x^{\frac{1}{x}}\) has a critical point when the derivative equals zero. Solving \(\frac{-\ln{x} + 1}{x^2} = 0\) gives \(x = e\), the base of natural logarithms. This critical point is crucial because it helps us decide how the function behaves and, in this case, how to compare \(e^\pi\) and \(\pi^e\).

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It's like asking, "If I change \(x\) slightly, how does \(f(x)\) change?". This makes derivatives essential for understanding the behavior of functions, especially for finding critical points.

• To find the derivative, apply differentiation rules such as the product, quotient, and chain rules.
• Implicit differentiation can be used when dealing with functions presented in terms of other functions, like in logarithmic cases.
• The result of differentiation gives us a function that represents the slope at any given point on the original function.

In our example, the function itself isn't directly simple to differentiate, hence the logarithmic transformation. By taking \( \ln(f(x)) \), we simplified the task. Calculating the derivative gave us insight into the function's growth and decline, pinpointing the critical value \(x=e\).

Logarithms

Logarithms are the inverse operation of exponentiation and aid in solving equations involving exponential terms by transforming multiplicative processes into additive ones. This property makes them invaluable for tackling complex algebraic equations.

• Using properties like \( \ln(a^b) = b\ln(a) \), we can simplify exponential terms.
• Applying logarithms can transform a function into a form that's easier to differentiate or integrate.

In the problem, we used \( \ln(x^\frac{1}{x}) = \frac{1}{x}\ln(x) \) to transform the original function. This logarithmic transformation allowed us to apply the differentiation rules more readily, showing the power of logarithms in dealing with exponential functions. It led to discovering where the function reaches its critical behavior at \(x = e\).

Exponential Functions

Exponential functions involve terms where a constant base is raised to a variable exponent, such as \(a^x\). These functions grow or decay at rates proportional to their current value, making them significant in various real-world scenarios ranging from population growth to compound interest.

• They often outpace other types of functions due to their rapid growth or decay.
• They are naturally linked to logarithms, as logarithms are the inverse of exponentiation.
• The base \(e\) is particularly important because it leads to the natural exponential function \(e^x\), which has unique properties that simplify calculus operations.

In the exercise, exponential functions are central to the comparison between \(e^\pi\) and \(\pi^e\). By understanding how the function \(f(x)=x^{\frac{1}{x}}\) behaves at its critical point, we conclude that \(e^{\pi} > \pi^{e}\). This exemplifies the power of exponential growth, even with different bases involved, underscoring the unique efficiency of the natural base \(e\).