Animated Solution for Mathematics - Differentiation: If the function $f(x)=(\frac{1}{x}{)}^{2x};x>0$ attains the maximum value at $x=\frac{1}{e}$ then :

Imagine you are standing at the edge of a mathematical cliff, staring into the abyss of two of the most famous numbers in the universe: $e$ and $\pi$. We often encounter them in isolation, but today, we are going to pit them against each other in a battle of exponents.

Which is larger: $e^{\pi }$ or ${\pi }^e$? This is not just a calculation; it is a journey into the heart of functional analysis.

The Setup and the Logarithmic Key

We start with the function $f(x)=(\frac{1}{x}{)}^{2x}$. The problem tells us that this function reaches its maximum at $x=\frac{1}{e}$.

When you see a variable in the base and the exponent, your first instinct should always be logarithmic differentiation. We take the natural log of both sides:

Since $\ln (\frac{1}{x})=-\ln x$, this simplifies beautifully to $\ln f(x)=-2x\ln x$. Differentiating this using the product rule, we get:

Setting this to zero gives us the critical point $x=\frac{1}{e}$, confirming the peak of our function.

The Bridge to the Comparison

Now, how does this help us compare $e^{\pi }$ and ${\pi }^e$? If we take the $e\pi$-th root of both expressions, we are essentially comparing $e^{\frac{1}{e}}$ and ${\pi }^{\frac{1}{\pi }}$.

This is the bridge! We define a new, auxiliary function $g(x)=x^{\frac{1}{x}}$. If we can determine whether $g(x)$ is increasing or decreasing, we can determine the relative sizes of $g(e)$ and $g(\pi )$.

The Analysis of Growth

Let us look at $g(x)=x^{\frac{1}{x}}$. Again, we use our trusty logarithmic differentiation:

Differentiating with respect to $x$ gives us:

Therefore, the derivative is:

This derivative is the key to everything. When $x=e$, the derivative is zero. When $x>e$, $\ln x>1$, which means $1-\ln x<0$. This tells us that for all $x>e$, the function $g(x)$ is strictly decreasing.

The Final Verdict

We know that $\pi \approx 3.14$ and $e\approx 2.718$, so $\pi >e$. Since $g(x)$ is strictly decreasing for $x>e$, it must be that $g(e)>g(\pi )$.

Substituting back, we get:

Now, we simply raise both sides to the power of $e\pi$. The left side becomes $(e^{\frac{1}{e}}{)}^{e\pi }=e^{\pi }$, and the right side becomes $({\pi }^{\frac{1}{\pi }}{)}^{e\pi }={\pi }^e$.

Thus, $e^{\pi }>{\pi }^e$. We have conquered the mountain! The elegance of this result lies in the fact that we did not need a calculator; we only needed the power of calculus to reveal the hidden behavior of these transcendental numbers.