Key Concepts and Formulas

• Differentiability: A function $g(x)$ is differentiable at a point $x=a$ if the limit of the difference quotient exists: $g'(a) = \lim_{h \to 0} \frac{g(a+h) - g(a)}{h}$. For a piecewise function, differentiability at the junction points requires continuity and equality of the left-hand and right-hand derivatives.
• Continuity: A function $g(x)$ is continuous at a point $x=a$ if $\lim_{x \to a^-} g(x) = \lim_{x \to a^+} g(x) = g(a)$.
• Finding the Maximum of a Function: To find the maximum of a function $f(t)$ over an interval $[0, x]$, we need to analyze its critical points (where $f'(t) = 0$ or is undefined) and the endpoints of the interval. The maximum will occur at one of these points.

Step-by-Step Solution

Step 1: Analyze the function for $0 \le x \le 3$. The function $g(x)$ for $0 \le x \le 3$ is defined as the maximum value of the polynomial $f(t) = t^3 - 6t^2 + 9t - 3$ for $0 \le t \le x$. Let's first find the critical points of $f(t)$ by taking its derivative: $f'(t) = 3t^2 - 12t + 9$ Setting $f'(t) = 0$ to find critical points: $3t^2 - 12t + 9 = 0$ $t^2 - 4t + 3 = 0$ $(t - 1)(t - 3) = 0$ The critical points are $t = 1$ and $t = 3$.

Now, let's evaluate $f(t)$ at the critical points and the endpoint $t=0$ to understand its behavior: $f(0) = 0^3 - 6(0)^2 + 9(0) - 3 = -3$ $f(1) = 1^3 - 6(1)^2 + 9(1) - 3 = 1 - 6 + 9 - 3 = 1$ $f(3) = 3^3 - 6(3)^2 + 9(3) - 3 = 27 - 54 + 27 - 3 = -3$

We can see that $f(t)$ has a local maximum at $t=1$ and a local minimum at $t=3$. The behavior of $f(t)$ on $[0, 3]$ is:

• From $t=0$ to $t=1$, $f'(t) > 0$, so $f(t)$ is increasing from $-3$ to $1$.
• From $t=1$ to $t=3$, $f'(t) < 0$, so $f(t)$ is decreasing from $1$ to $-3$.

Step 2: Determine $g(x)$ for $0 \le x \le 3$. Now, $g(x) = \mathop {\max }\limits_{0 \le t \le x} \{f(t)\}$.

• For $0 \le x \le 1$: Since $f(t)$ is increasing on $[0, x]$ in this interval, the maximum value of $f(t)$ on $[0, x]$ is $f(x)$. So, $g(x) = f(x) = x^3 - 6x^2 + 9x - 3$.
• For $1 < x \le 3$: The maximum value of $f(t)$ on $[0, x]$ occurs at $t=1$ (the local maximum). The value is $f(1) = 1$. So, $g(x) = f(1) = 1$.

Therefore, for $0 \le x \le 3$, $g(x)$ is: g(x) = \left\{ {\matrix{ {x^3} - 6{x^2} + 9x - 3,} & {0 \le x \le 1} \cr {1,} & {1 < x \le 3} \cr } } \right.

Step 3: Analyze the function for $3 < x \le 4$. For $3 < x \le 4$, $g(x) = 4 - x$. This is a linear function.

Step 4: Combine the definitions of $g(x)$ and check for continuity. The complete definition of $g(x)$ is: g(x) = \left\{ {\matrix{ {x^3} - 6{x^2} + 9x - 3,} & {0 \le x \le 1} \cr {1,} & {1 < x \le 3} \cr {4 - x,} & {3 < x \le 4} \cr } } \right.

We need to check for continuity at the junction points $x=1$ and $x=3$.

• At $x=1$:

  • Left-hand limit: $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} (x^3 - 6x^2 + 9x - 3) = 1^3 - 6(1)^2 + 9(1) - 3 = 1 - 6 + 9 - 3 = 1$.
  • Right-hand limit: $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} 1 = 1$.
  • Function value: $g(1) = 1^3 - 6(1)^2 + 9(1) - 3 = 1$. Since $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^+} g(x) = g(1)$, $g(x)$ is continuous at $x=1$.

• At $x=3$:

  • Left-hand limit: $\lim_{x \to 3^-} g(x) = \lim_{x \to 3^-} 1 = 1$.
  • Right-hand limit: $\lim_{x \to 3^+} g(x) = \lim_{x \to 3^+} (4 - x) = 4 - 3 = 1$.
  • Function value: $g(3) = 1$. Since $\lim_{x \to 3^-} g(x) = \lim_{x \to 3^+} g(x) = g(3)$, $g(x)$ is continuous at $x=3$.

Step 5: Check for differentiability at the junction points and within the intervals. We need to find the derivative of $g(x)$ in each interval.

• For $0 < x < 1$: $g'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x - 3) = 3x^2 - 12x + 9$.
• For $1 < x < 3$: $g'(x) = \frac{d}{dx}(1) = 0$.
• For $3 < x < 4$: $g'(x) = \frac{d}{dx}(4 - x) = -1$.

Now, let's check differentiability at $x=1$ and $x=3$.

• At $x=1$:

  • Left-hand derivative: $\lim_{x \to 1^-} g'(x) = \lim_{x \to 1^-} (3x^2 - 12x + 9) = 3(1)^2 - 12(1) + 9 = 3 - 12 + 9 = 0$.
  • Right-hand derivative: $\lim_{x \to 1^+} g'(x) = \lim_{x \to 1^+} 0 = 0$. Since the left-hand and right-hand derivatives are equal, $g(x)$ is differentiable at $x=1$.

• At $x=3$:

  • Left-hand derivative: $\lim_{x \to 3^-} g'(x) = \lim_{x \to 3^-} 0 = 0$.
  • Right-hand derivative: $\lim_{x \to 3^+} g'(x) = \lim_{x \to 3^+} (-1) = -1$. Since the left-hand derivative ($0$) and the right-hand derivative ($-1$) are not equal, $g(x)$ is NOT differentiable at $x=3$.

Step 6: Check for differentiability within the open intervals.

• For $0 < x < 1$, $g'(x) = 3x^2 - 12x + 9$. This is a polynomial and is differentiable everywhere.
• For $1 < x < 3$, $g'(x) = 0$. This is a constant and is differentiable everywhere.
• For $3 < x < 4$, $g'(x) = -1$. This is a constant and is differentiable everywhere.

The only point in the interval $(0, 4)$ where $g(x)$ is not differentiable is at $x=3$.

Step 7: Consider endpoints and the interval (0, 4). The question asks for the number of points in the interval (0, 4) where $g(x)$ is NOT differentiable. We have analyzed the junction points $x=1$ and $x=3$.

• At $x=1$, $g(x)$ is differentiable.
• At $x=3$, $g(x)$ is not differentiable.

The function is differentiable within the open intervals $(0, 1)$, $(1, 3)$, and $(3, 4)$.

Therefore, the only point in $(0, 4)$ where $g(x)$ is not differentiable is $x=3$.

Common Mistakes & Tips

• Confusing Maximum Function with the Polynomial: Remember that $g(x)$ is the maximum value of the polynomial $f(t)$ up to $x$. This means $g(x)$ might not be equal to $f(x)$ for all $x$.
• Forgetting to Check Continuity Before Differentiability: Differentiability at a point implies continuity at that point. Always check continuity at junction points first. If a function is not continuous at a point, it cannot be differentiable there.
• Incorrectly Evaluating the Maximum: When dealing with $\max_{0 \le t \le x} f(t)$, consider the critical points of $f(t)$ and how they affect the maximum value as $x$ changes.

Summary

We defined the function $g(x)$ by first analyzing the behavior of the polynomial $f(t) = t^3 - 6t^2 + 9t - 3$ and finding its maximum over the interval $[0, x]$. This led to a piecewise definition of $g(x)$ for $0 \le x \le 3$. We then combined this with the definition for $3 < x \le 4$ to get the complete piecewise function. We checked for continuity at the junction points $x=1$ and $x=3$ and found that $g(x)$ is continuous everywhere in $[0, 4]$. Next, we calculated the derivatives in each interval and checked for differentiability at the junction points. We found that $g(x)$ is differentiable at $x=1$ but not at $x=3$ because the left-hand and right-hand derivatives do not match. The function is differentiable within the open intervals. Thus, there is only one point in $(0, 4)$ where $g(x)$ is not differentiable.

The final answer is $\boxed{3}$.