Question:

Step 1: Understand the objective and the provided context. The problem asks for a comparison between $e^\pi$ and $\pi^e$. The initial statement about the function $f(x)=\left(\dfrac{1}{x}\right)^{2x}$ attaining its maximum at $x=\dfrac{1}{e}$ serves as a setup, indicating that properties related to the number $e$ and exponential functions are central to the problem. To compare $e^\pi$ and $\pi^e$, we can analyze the behavior of the function $x^{1/x}$. Step 2: Define an auxiliary function for comparison. To compare quantities of the form $a^b$ and $b^a$, it is often useful to analyze the function $g(x) = x^{1/x}$ for $x>0$. Comparing $e^\pi$ and $\pi^e$ is equivalent to comparing $e^{1/e}$ and $\pi^{1/\pi}$ by raising both to the power of $e\pi$. Step 3: Analyze the auxiliary function $g(x) = x^{1/x}$. To understand the behavior of $g(x)$, we take its natural logarithm and then differentiate. Let $g(x) = x^{1/x}$. Taking the natural logarithm of both sides: $$ \ln(g(x)) = \ln(x^{1/x}) = \frac{1}{x} \ln x $$ Now, differentiate $\ln(g(x))$ with respect to $x$ using the quotient rule: $$ \frac{g'(x)}{g(x)} = \frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{\left(\frac{1}{x}\right) \cdot x - (\ln x) \cdot 1}{x^2} = \frac{1 - \ln x}{x^2} $$ Multiplying by $g(x)$, we find the derivative of $g(x)$: $$ g'(x) = g(x) \cdot \frac{1 - \ln x}{x^2} = x^{1/x} \frac{1 - \ln x}{x^2} $$ Step 4: Determine the critical point and intervals of increase/decrease for $g(x)$. To find critical points, we set $g'(x) = 0$: $$ x^{1/x} \frac{1 - \ln x}{x^2} = 0 $$ Since $x^{1/x} > 0$ and $x^2 > 0$ for $x>0$, the only way for $g'(x)$ to be zero is if $1 - \ln x = 0$: $$ \ln x = 1 \implies x = e $$ This indicates that $x=e$ is a critical point. Now, let's analyze the sign of $g'(x)$ around $x=e$: For $x \in (0, e)$, $\ln x < 1$, so $1 - \ln x > 0$. Therefore, $g'(x) > 0$, meaning $g(x)$ is increasing. For $x \in (e, \infty)$, $\ln x > 1$, so $1 - \ln x < 0$. Therefore, $g'(x) < 0$, meaning $g(x)$ is decreasing. This confirms that $g(x)$ attains its maximum value at $x=e$. Step 5: Compare $e^{1/e}$ and $\pi^{1/\pi}$ using the behavior of $g(x)$. We know that $g(x)$ increases up to $x=e$ and then decreases for $x>e$. The value of $e \approx 2.718$ and $\pi \approx 3.141$. Since $e < \pi$, and both $e$ and $\pi$ are in the domain where $g(x)$ is defined, and $g(x)$ is decreasing for all $x > e$, it follows that $g(e)$ must be greater than $g(\pi)$. Thus, we have: $$ e^{1/e} > \pi^{1/\pi} $$ Step 6: Transform the inequality to compare $e^\pi$ and $\pi^e$. To compare $e^\pi$ and $\pi^e$, we raise both sides of the inequality $e^{1/e} > \pi^{1/\pi}$ to the power of $e\pi$. Since $e\pi$ is a positive number, the direction of the inequality remains unchanged. $$ \left(e^{1/e}\right)^{e\pi} > \left(\pi^{1/\pi}\right)^{e\pi} $$ Using the exponent rule $(a^m)^n = a^{mn}$: $$ e^{\frac{1}{e} \cdot e\pi} > \pi^{\frac{1}{\pi} \cdot e\pi} $$ $$ e^{\pi} > \pi^{e} $$ Step 7: Conclude the final answer. Based on our analysis, we find that $e^\pi > \pi^e$. Comparing this result with the given options: Option 1: $e^\pi < \pi^e$ Option 2: $e^\pi > \pi^e$ Option 3: $(2e)^\pi > \pi^{2e}$ Option 4: $e^{2\pi} < (2\pi)^e$ The correct option is Option 2. The final answer is $\boxed{\text{e^\pi > \pi^e}}$.