[b](a)[/b] Sketch the diagram of the function $f$ if
\[f(x)=4x(1-|x|) , \quad |x| \leq 1.\]

[b](b)[/b] Does there exist derivative of $f$ in the point $x=0 \ ?$

[b](c)[/b] Let $g$ be a function such that
\[g(x)=\left\{\begin{array}{cc}\frac{f(x)}{x} \quad :  x \neq 0\\ \text{ } \\ 4 \ \ \ \ \quad : x=0\end{array}\right.\]
Is the function $g$ continuous in the point $x=0 \ ?$ 

[b](d)[/b] Sketch the diagram of $g.$
**(a)** Sketch the diagram of the function \( f \) if
\[ f(x) = 4x(1 - |x|), \quad |x| \leq 1. \]

1. The function \( f(x) \) is defined piecewise for \( |x| \leq 1 \). We can rewrite it as:
   \[
   f(x) = \begin{cases}
   4x(1 - x), & 0 \leq x \leq 1 \\
   4x(1 + x), & -1 \leq x < 0
   \end{cases}
   \]

2. For \( 0 \leq x \leq 1 \):
   \[
   f(x) = 4x(1 - x) = 4x - 4x^2
   \]
   This is a downward-opening parabola.

3. For \( -1 \leq x < 0 \):
   \[
   f(x) = 4x(1 + x) = 4x + 4x^2
   \]
   This is an upward-opening parabola.

4. The function \( f(x) \) is continuous at \( x = 0 \) since both pieces meet at \( f(0) = 0 \).

**(b)** Does there exist a derivative of \( f \) at the point \( x = 0 \)?

1. To check the existence of the derivative at \( x = 0 \), we need to compute the left-hand and right-hand derivatives.

2. Right-hand derivative at \( x = 0 \):
   \[
   \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{4x - 4x^2 - 0}{x} = \lim_{x \to 0^+} (4 - 4x) = 4
   \]

3. Left-hand derivative at \( x = 0 \):
   \[
   \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{4x + 4x^2 - 0}{x} = \lim_{x \to 0^-} (4 + 4x) = 4
   \]

4. Since both the left-hand and right-hand derivatives are equal, the derivative of \( f \) at \( x = 0 \) exists and is equal to 4.

**(c)** Let \( g \) be a function such that
\[ g(x) = \begin{cases}
\frac{f(x)}{x}, & x \neq 0 \\
4, & x = 0
\end{cases} \]
Is the function \( g \) continuous at the point \( x = 0 \)?

1. To check the continuity of \( g \) at \( x = 0 \), we need to verify if \( \lim_{x \to 0} g(x) = g(0) \).

2. For \( x \neq 0 \):
   \[
   g(x) = \frac{f(x)}{x}
   \]

3. As \( x \to 0 \):
   \[
   \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \begin{cases}
   4 - 4x, & 0 < x \leq 1 \\
   4 + 4x, & -1 \leq x < 0
   \end{cases} = 4
   \]

4. Since \( g(0) = 4 \) and \( \lim_{x \to 0} g(x) = 4 \), the function \( g \) is continuous at \( x = 0 \).

**(d)** Sketch the diagram of \( g \).

1. For \( x \neq 0 \):
   \[
   g(x) = \frac{f(x)}{x} = \begin{cases}
   4 - 4x, & 0 < x \leq 1 \\
   4 + 4x, & -1 \leq x < 0
   \end{cases}
   \]

2. At \( x = 0 \):
   \[
   g(0) = 4
   \]

3. The function \( g(x) \) is a piecewise linear function with a value of 4 at \( x = 0 \).

The final answer is \( \boxed{4} \)