To solve this problem, let's break it down into two parts:

Part 1: Show that $f(0) = 0$

We are given that for all $x$, the inequality $|f(x)| \leq x^2$ holds. Specifically, we can substitute $x = 0$ into this inequality:

$|f(0)| \leq 0^2 = 0$

This implies:

$|f(0)| \leq 0$

Since the absolute value of any real number is non-negative, $|f(0)| \geq 0$. The only way for $|f(0)|$ to satisfy both $|f(0)| \leq 0$ and $|f(0)| \geq 0$ is if $f(0) = 0$.

Thus, we have shown that: $f(0) = 0$

Part 2: Show that $f'(0) = 0$

The derivative $f'(0)$ is defined as the limit:

$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$

Since we already know $f(0) = 0$, this simplifies to:

$f'(0) = \lim_{h \to 0} \frac{f(h)}{h}$

From the given condition $|f(x)| \leq x^2$, we have:

$|f(h)| \leq h^2$

Dividing both sides by $|h|$ (for $h \neq 0$):

$\frac{|f(h)|}{|h|} \leq \frac{h^2}{|h|} = |h|$

Thus, we have:

$\left| \frac{f(h)}{h} \right| \leq |h|$

Taking the limit as $h \to 0$, the right-hand side tends to 0, meaning:

$\lim_{h \to 0} \left| \frac{f(h)}{h} \right| = 0$

Since the absolute value tends to 0, the limit itself must also tend to 0:

$\lim_{h \to 0} \frac{f(h)}{h} = 0$

Thus, $f'(0) = 0$.

Conclusion:

• We have shown that $f(0) = 0$.
• We have also shown that $f'(0) = 0$.

Would you like further clarification on any of the steps?

Here are 5 related questions to think about:

1. What does the derivative of a function represent geometrically at a point?
2. How does the condition $|f(x)| \leq x^2$ restrict the growth of $f(x)$ near $x = 0$?
3. Can we deduce more properties about $f(x)$ from the inequality $|f(x)| \leq x^2$?
4. If the inequality were $|f(x)| \leq x^3$, what would change in the argument for $f'(0) = 0$?
5. How would you interpret $f(0) = 0$ and $f'(0) = 0$ in terms of limits?

Tip: When calculating derivatives at a point, always consider how the behavior of the function near that point influences the derivative.