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Let's break down each part of the problem and solve it using the definition of the derivative.

### 1. Showing that \( f(x) = x |x| \) has a derivative at \( x = 0 \)
We first define the function \( f(x) = x |x| \). This can be rewritten in a piecewise form to handle the absolute value:

\[
f(x) = 
\begin{cases} 
x^2 & \text{if } x \geq 0, \\
-x^2 & \text{if } x < 0.
\end{cases}
\]

We now apply the definition of the derivative at \( x = 0 \):
\[
f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}.
\]

We know that \( f(0) = 0 \), so the derivative becomes:
\[
f'(0) = \lim_{h \to 0} \frac{f(h)}{h}.
\]

Let's check the limit from both sides:

- **For \( h > 0 \):** When \( h > 0 \), \( f(h) = h^2 \). So,
  \[
  f'(0) = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} h = 0.
  \]

- **For \( h < 0 \):** When \( h < 0 \), \( f(h) = -h^2 \). So,
  \[
  f'(0) = \lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} -h = 0.
  \]

Since the limit from both sides gives the same value, we conclude that \( f'(0) = 0 \), meaning that **the function has a derivative at \( x = 0 \)**.

### 2. Showing that \( f(x) = |2x - 2| \) does not have a derivative at \( x = 1 \)

We are given the function \( f(x) = |2x - 2| \). To show that it does not have a derivative at \( x = 1 \), let's first rewrite it in a piecewise form based on the definition of absolute value:

\[
f(x) = 
\begin{cases} 
2x - 2 & \text{if } x \geq 1, \\
-(2x - 2) & \text{if } x < 1.
\end{cases}
\]

Now, we apply the definition of the derivative at \( x = 1 \):
\[
f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}.
\]

We know that \( f(1) = 0 \), so the derivative becomes:
\[
f'(1) = \lim_{h \to 0} \frac{f(1+h)}{h}.
\]

Let's check the limit from both sides:

- **For \( h > 0 \):** When \( h > 0 \), \( f(1+h) = 2(1+h) - 2 = 2h \). So,
  \[
  f'(1) = \lim_{h \to 0^+} \frac{2h}{h} = 2.
  \]

- **For \( h < 0 \):** When \( h < 0 \), \( f(1+h) = -(2(1+h) - 2) = -2h \). So,
  \[
  f'(1) = \lim_{h \to 0^-} \frac{-2h}{h} = -2.
  \]

Since the limit from the right-hand side is 2 and from the left-hand side is -2, the derivative does not exist at \( x = 1 \). Hence, **the function does not have a derivative at \( x = 1 \)**.

---

Would you like more details on these steps or explanations? Here are some additional related questions for you to consider:

1. How is the definition of the derivative related to limits?
2. What happens to a function's continuity when its derivative does not exist at a point?
3. Why does the absolute value function often cause issues with differentiability?
4. How does the piecewise definition of functions affect derivative calculations?
5. Can you identify other points where functions involving absolute values are non-differentiable?

**Tip:** Always check both the left-hand and right-hand limits when determining if a function has a derivative at a given point.

Using the definition of the derivative, show that: 1. The function f(x) = x|x| has a derivative at x = 0. 2. The function f(x) = |2x - 2| does not have a derivative at x = 1.

Explore how to determine if the function f(x) = x|x| has a derivative at x = 0 and if f(x) = |2x - 2| does not have a derivative at x = 1 using the definition of derivatives. Detailed breakdown of piecewise functions and absolute values.

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