Suppose f is a function with the property that |f(x)| is less than or equal to x squared for all x. Show that f(0)=0 and show that f'(0)=0

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.

Suppose f is a function with the property that |f(x)| is less than or equal to x squared for all x. Show that f(0)=0 and show that f'(0)=0.

Explore how to prove that f(0)=0 and f'(0)=0 for a function bounded by |f(x)| <= x^2 using limit definitions of derivatives and the squeeze theorem. This solution delves into key calculus concepts like absolute values and derivatives, suitable for Grades 11-12 and early college students.

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