Key Concepts and Formulas

• Definition of the Derivative: The derivative of a function $f(x)$ at a point $x$ is defined as: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
• Properties of Absolute Value: For any real numbers $a$ and $b$, $|a \cdot b| = |a| \cdot |b|$ and $|a/b| = |a|/|b|$ (if $b \neq 0$). Also, if $|a| \le b$, then $-b \le a \le b$.
• Consequences of a Derivative Being Zero: If $f'(x) = 0$ for all $x$ in an interval, then $f(x)$ is a constant function on that interval.

Step-by-Step Solution

Step 1: Use the definition of the derivative to analyze $f'(x)$. We are given that $f$ is a real-valued differentiable function. The definition of the derivative is: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ To understand the behavior of $f'(x)$, we will examine its absolute value.

Step 2: Apply the given inequality to the derivative. We are given the inequality $|f(x) - f(y)| \le (x-y)^2$ for all $x, y \in \mathbb{R}$. Let's substitute $y = x+h$ into this inequality. Then $x-y = x - (x+h) = -h$, and $(x-y)^2 = (-h)^2 = h^2$. The inequality becomes: $|f(x) - f(x+h)| \le h^2$ Since $|a-b| = |b-a|$, we have $|f(x+h) - f(x)| \le h^2$.

Now, let's consider the term inside the limit for the derivative: $\left| \frac{f(x+h) - f(x)}{h} \right| = \frac{|f(x+h) - f(x)|}{|h|}$ Using the inequality derived from the problem statement, we have: $\frac{|f(x+h) - f(x)|}{|h|} \le \frac{h^2}{|h|}$

Step 3: Evaluate the limit of the absolute value of the difference quotient. Now we take the limit as $h \to 0$: $|f'(x)| = \left| \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \right|$ Since the absolute value function is continuous, we can move the limit inside the absolute value: $|f'(x)| = \lim_{h \to 0} \left| \frac{f(x+h) - f(x)}{h} \right|$ From Step 2, we know that $\left| \frac{f(x+h) - f(x)}{h} \right| \le \frac{h^2}{|h|}$. Therefore, $|f'(x)| \le \lim_{h \to 0} \frac{h^2}{|h|}$ Let's evaluate the limit on the right side: $\lim_{h \to 0} \frac{h^2}{|h|}$ If $h > 0$, $|h| = h$, so $\frac{h^2}{h} = h$. The limit as $h \to 0^+$ is $\lim_{h \to 0^+} h = 0$. If $h < 0$, $|h| = -h$, so $\frac{h^2}{-h} = -h$. The limit as $h \to 0^-$ is $\lim_{h \to 0^-} (-h) = 0$. Since the left-hand limit and the right-hand limit are equal, the limit exists and is 0. So, we have: $|f'(x)| \le 0$

Step 4: Deduce the value of $f'(x)$. We have established that $|f'(x)| \le 0$. The absolute value of any real number is always non-negative ($|a| \ge 0$). The only way for $|f'(x)|$ to be less than or equal to 0 is if $|f'(x)|$ is exactly 0. Therefore, $|f'(x)| = 0$, which implies $f'(x) = 0$ for all $x \in \mathbb{R}$.

Step 5: Determine the function $f(x)$. Since $f'(x) = 0$ for all $x$, the function $f(x)$ must be a constant function. Let $f(x) = C$, where $C$ is a constant.

Step 6: Use the initial condition to find the constant. We are given that $f(0) = 0$. Since $f(x) = C$ for all $x$, we can substitute $x=0$: $f(0) = C$ Given $f(0) = 0$, we have $C = 0$. Therefore, the function $f(x)$ is $f(x) = 0$ for all $x \in \mathbb{R}$.

Step 7: Calculate $f(1)$. Since $f(x) = 0$ for all $x$, we can find $f(1)$ by substituting $x=1$: $f(1) = 0$.

Common Mistakes & Tips

• Incorrectly handling the absolute value: Be careful when taking the limit of an expression involving absolute values. Ensure that the inequality derived from the problem statement is correctly applied to the difference quotient.
• Assuming $f'(x)$ is directly $(x-y)^2/h$: The inequality $|f(x) - f(y)| \le (x-y)^2$ is a condition on the function, not a direct formula for its derivative. You must use the definition of the derivative and apply the inequality within the limit.
• Forgetting the initial condition: The initial condition $f(0)=0$ is crucial for determining the specific constant of integration. Without it, we would only know that $f(x)$ is a constant, but not its value.

Summary

The problem provides a constraint on a differentiable function $f(x)$: $|f(x) - f(y)| \le (x-y)^2$. By using the definition of the derivative and applying this inequality, we showed that $|f'(x)| \le 0$. This implies that $f'(x) = 0$ for all $x$. A function with a derivative of zero everywhere must be a constant function. Using the given initial condition $f(0)=0$, we determined that the constant is 0, meaning $f(x) = 0$ for all $x$. Consequently, $f(1) = 0$.

The final answer is $\boxed{0}$.