Question:

Step 1: Determine the function $f(x)$. The problem states that $f:(0,\infty) \to \mathbb{R}$ is a differentiable function satisfying $f(1)=0$ and $f'(1)=1$. The given functional equation is $2f(x) = f(x) + f\left(\dfrac{x}{y}\right)$ for all $x, y > 0$. This equation simplifies to $f(x) = f\left(\dfrac{x}{y}\right)$. If $f(x) = f\left(\dfrac{x}{y}\right)$ for all $x,y > 0$, then setting $x=1$ yields $f(1) = f\left(\dfrac{1}{y}\right)$. Since $f(1)=0$, this implies $f(z)=0$ for all $z>0$. However, if $f(x)=0$ for all $x$, then $f'(x)=0$, which contradicts the given condition $f'(1)=1$. This indicates a likely typo in the functional equation provided in the problem statement. Given the conditions $f(1)=0$ and $f'(1)=1$, the function $f(x) = \ln x$ is the unique solution to the functional equation $f(xy) = f(x) + f(y)$ (Cauchy's functional equation for logarithms) that satisfies these initial conditions. Let us verify $f(x) = \ln x$ with the given conditions: 1. $f(1) = \ln 1 = 0$. This condition is satisfied. 2. $f'(x) = \dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$. So, $f'(1) = \dfrac{1}{1} = 1$. This condition is also satisfied. Therefore, we proceed with $f(x) = \ln x$. Step 2: Analyze the properties of $f(x)$. * **Local extrema of $f(x)$:** The derivative of $f(x)$ is $f'(x) = \dfrac{1}{x}$. For $x > 0$, $f'(x) > 0$. Since $f'(x) > 0$ for all $x$ in the domain $(0, \infty)$, $f(x)$ is strictly increasing. A strictly increasing function has no local maxima and no local minima. * **Limit evaluation:** Consider the limit $\displaystyle\lim_{x\to 0^+} \left[\dfrac{f(x+1)}{x}\right]$. $$ \lim_{x\to 0^+} \dfrac{f(x+1)}{x} = \lim_{x\to 0^+} \dfrac{\ln(x+1)}{x} $$ This is a standard limit, which evaluates to $1$. For $x > 0$ and sufficiently small, it is known that $\ln(1+x) < x$. Therefore, $\dfrac{\ln(1+x)}{x} < 1$. Thus, as $x \to 0^+$, $\dfrac{\ln(x+1)}{x}$ approaches $1$ from values less than $1$. Let $L(x) = \dfrac{\ln(x+1)}{x}$. Then $\displaystyle\lim_{x\to 0^+} [L(x)] = [1^-] = 0$. * **Roots of $f(x) = ex$:** Consider the equation $f(x) = ex$, which is $\ln x = ex$. Let $g(x) = \ln x - ex$. We seek the roots of $g(x) = 0$. The derivative is $g'(x) = \dfrac{1}{x} - e$. Setting $g'(x) = 0$ yields $\dfrac{1}{x} = e$, so $x = \dfrac{1}{e}$. The second derivative is $g''(x) = -\dfrac{1}{x^2}$. For $x > 0$, $g''(x) < 0$, indicating that $x = \dfrac{1}{e}$ is a local maximum. The maximum value of $g(x)$ is $g\left(\dfrac{1}{e}\right) = \ln\left(\dfrac{1}{e}\right) - e\left(\dfrac{1}{e}\right) = -1 - 1 = -2$. Since the maximum value of $g(x)$ is $-2$, which is less than $0$, $g(x) < 0$ for all $x > 0$. Therefore, the equation $\ln x = ex$ has no roots. * **Solutions to $2e \cdot f(x) = x$:** Consider the equation $2e \cdot f(x) = x$, which is $2e \ln x = x$. Let $h(x) = 2e \ln x - x$. We seek the roots of $h(x) = 0$. The derivative is $h'(x) = \dfrac{2e}{x} - 1$. Setting $h'(x) = 0$ yields $\dfrac{2e}{x} = 1$, so $x = 2e$. The second derivative is $h''(x) = -\dfrac{2e}{x^2}$. For $x > 0$, $h''(x) < 0$, indicating that $x = 2e$ is a local maximum. The maximum value of $h(x)$ is $h(2e) = 2e \ln(2e) - 2e = 2e(\ln 2 + \ln e) - 2e = 2e(\ln 2 + 1) - 2e = 2e \ln 2$. Since $\ln 2 > 0$, $h(2e) > 0$. As $x \to 0^+$, $h(x) = 2e \ln x - x \to -\infty$. As $x \to \infty$, $h(x) = 2e \ln x - x$. Since $x$ grows faster than $\ln x$, $h(x) \to -\infty$. Since $h(x)$ starts at $-\infty$, increases to a positive maximum $h(2e) = 2e \ln 2$, and then decreases back to $-\infty$, by the Intermediate Value Theorem, there must be exactly two distinct roots for $h(x) = 0$.