Key Concepts and Formulas

• Differentiability of Piecewise Functions: A function $f(x)$ is differentiable at a point $x=a$ if it is continuous at $x=a$ and the left-hand derivative (LHD) equals the right-hand derivative (RHD) at $x=a$.
  • Continuity at $x=a$: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.
  • LHD at $x=a$: $f'_-(a) = \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$.
  • RHD at $x=a$: $f'_+(a) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$.
• Properties of Greatest Integer Function: $[t]$ is the greatest integer less than or equal to $t$. It is discontinuous at integer values.
• Definite Integration of Piecewise Functions: The integral of a piecewise function over an interval is the sum of the integrals of each piece over its respective subinterval.

Step-by-Step Solution

Part 1: Analyzing Differentiability

The function $f(x)$ is defined in pieces. We need to check for differentiability at the points where the definition changes: $x=2$, $x=3$, and $x=5$.

Step 1: Analyze the first piece $f(x) = \max \{t^3 - 3t\} \text{ for } t \le x, \text{ where } x \le 2$. Let $g(t) = t^3 - 3t$. To find $\max \{t^3 - 3t\} \text{ for } t \le x$, we need to analyze the behavior of $g(t)$. $g'(t) = 3t^2 - 3 = 3(t^2 - 1)$. Critical points for $g(t)$ are $t=1$ and $t=-1$. $g''(t) = 6t$. $g''(-1) = -6 < 0$, so $t=-1$ is a local maximum, $g(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$. $g''(1) = 6 > 0$, so $t=1$ is a local minimum, $g(1) = (1)^3 - 3(1) = 1 - 3 = -2$.

For $t \le x$, the maximum value of $g(t)$ depends on $x$. If $x \le -1$, the maximum is achieved at $t=x$, so $f(x) = x^3 - 3x$. If $-1 < x \le 1$, the maximum is achieved at $t=-1$, so $f(x) = g(-1) = 2$. If $1 < x \le 2$, the maximum is achieved at $t=-1$, so $f(x) = g(-1) = 2$.

So, for $x \le 2$, the function $f(x)$ can be described as: f(x) = \left\{ {\matrix{ {x^3 - 3x} & ; & {x \le -1} \cr {2} & ; & {-1 < x \le 2} \cr } } \right.

Step 2: Check differentiability at $x=2$. We need to consider the definition for $x \le 2$ and $2 < x < 3$. From the first piece ($x \le 2$), for $x$ approaching 2 from the left, $f(x) = 2$. $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} 2 = 2$. $f(2) = 2$.

From the second piece ($2 < x < 3$), $f(x) = x^2 + 2x - 6$. $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 2x - 6) = 2^2 + 2(2) - 6 = 4 + 4 - 6 = 2$.

Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 2$, the function is continuous at $x=2$.

Now, let's check differentiability at $x=2$. LHD at $x=2$: For $x < 2$ and close to 2, $f(x) = 2$. $f'_-(2) = \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0^-} \frac{2 - 2}{h} = 0$.

RHD at $x=2$: For $x > 2$ and close to 2, $f(x) = x^2 + 2x - 6$. $f'(x) = 2x + 2$. $f'_+(2) = \lim_{x \to 2^+} f'(x) = 2(2) + 2 = 6$.

Since $f'_-(2) = 0 \neq f'_+(2) = 6$, the function is not differentiable at $x=2$.

Step 3: Check differentiability at $x=3$. We need to consider the definition for $2 < x < 3$ and $3 \le x \le 5$. From the second piece ($2 < x < 3$), $f(x) = x^2 + 2x - 6$. $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^2 + 2x - 6) = 3^2 + 2(3) - 6 = 9 + 6 - 6 = 9$.

From the third piece ($3 \le x \le 5$), $f(x) = [x - 3] + 9$. $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} ([x - 3] + 9) = [3^+ - 3] + 9 = [0^+] + 9 = 0 + 9 = 9$. $f(3) = [3 - 3] + 9 = [0] + 9 = 0 + 9 = 9$.

Since $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) = 9$, the function is continuous at $x=3$.

Now, let's check differentiability at $x=3$. LHD at $x=3$: For $x < 3$ and close to 3, $f(x) = x^2 + 2x - 6$. $f'(x) = 2x + 2$. $f'_-(3) = \lim_{x \to 3^-} f'(x) = 2(3) + 2 = 8$.

RHD at $x=3$: For $x > 3$ and close to 3, $f(x) = [x - 3] + 9$. For $3 < x < 4$, $[x-3] = 0$. So $f(x) = 0 + 9 = 9$. $f'_+(3) = \lim_{h \to 0^+} \frac{f(3+h) - f(3)}{h} = \lim_{h \to 0^+} \frac{([3+h - 3] + 9) - 9}{h} = \lim_{h \to 0^+} \frac{[h] + 9 - 9}{h} = \lim_{h \to 0^+} \frac{0}{h} = 0$.

Since $f'_-(3) = 8 \neq f'_+(3) = 0$, the function is not differentiable at $x=3$.

Step 4: Check differentiability at $x=5$. We need to consider the definition for $3 \le x \le 5$ and $x > 5$. From the third piece ($3 \le x \le 5$), $f(x) = [x - 3] + 9$. For $x$ approaching 5 from the left, we consider values like $x=4.9$. If $4 \le x < 5$, then $1 \le x-3 < 2$, so $[x-3] = 1$. Thus, $f(x) = 1 + 9 = 10$. $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} ([x - 3] + 9) = 1 + 9 = 10$. $f(5) = [5 - 3] + 9 = [2] + 9 = 2 + 9 = 11$.

Since $\lim_{x \to 5^-} f(x) = 10 \neq f(5) = 11$, the function is not continuous at $x=5$. Therefore, the function is not differentiable at $x=5$.

Step 5: Identify points of non-differentiability. The points where $f(x)$ is not differentiable are $x=2$, $x=3$, and $x=5$. The number of points where $f$ is not differentiable is $m=3$.

Part 2: Calculating the Definite Integral

We need to calculate $I = \int_{-2}^2 f(x) \, dx$. We need to use the definition of $f(x)$ for the interval $[-2, 2]$. From Step 1, we found that for $x \le 2$, f(x) = \left\{ {\matrix{ {x^3 - 3x} & ; & {x \le -1} \cr {2} & ; & {-1 < x \le 2} \cr } } \right.

So, the integral can be split into two parts: $I = \int_{-2}^{-1} (x^3 - 3x) \, dx + \int_{-1}^2 2 \, dx$

Step 6: Evaluate the first integral. $\int_{-2}^{-1} (x^3 - 3x) \, dx = \left[ \frac{x^4}{4} - \frac{3x^2}{2} \right]_{-2}^{-1}$ $= \left( \frac{(-1)^4}{4} - \frac{3(-1)^2}{2} \right) - \left( \frac{(-2)^4}{4} - \frac{3(-2)^2}{2} \right)$ $= \left( \frac{1}{4} - \frac{3}{2} \right) - \left( \frac{16}{4} - \frac{12}{2} \right)$ $= \left( \frac{1}{4} - \frac{6}{4} \right) - (4 - 6)$ $= \left( -\frac{5}{4} \right) - (-2)$ $= -\frac{5}{4} + 2 = -\frac{5}{4} + \frac{8}{4} = \frac{3}{4}$

Step 7: Evaluate the second integral. $\int_{-1}^2 2 \, dx = [2x]_{-1}^2$ $= 2(2) - 2(-1)$ $= 4 - (-2)$ $= 4 + 2 = 6$

Step 8: Calculate the total integral. $I = \frac{3}{4} + 6 = \frac{3}{4} + \frac{24}{4} = \frac{27}{4}$

Step 9: Form the ordered pair (m, I). We found $m=3$ and $I = \frac{27}{4}$. The ordered pair is $\left( 3, \frac{27}{4} \right)$.

Common Mistakes & Tips

• Incorrectly determining the maximum for the first piece: Be careful when finding the maximum of $t^3 - 3t$ for $t \le x$. Analyze the local extrema and how they affect the maximum value as $x$ changes.
• Errors in continuity and differentiability checks: Ensure you correctly apply the definitions of continuity (matching limits and function value) and differentiability (matching LHD and RHD). Pay close attention to the behavior of the greatest integer function near integers.
• Integration limits for piecewise functions: When integrating, correctly split the integral at the points where the function definition changes and ensure the correct piece of the function is used for each subinterval.

Summary

We first analyzed the differentiability of the function by examining its continuity and the equality of left-hand and right-hand derivatives at the critical points $x=2$, $x=3$, and $x=5$. We found that the function is not differentiable at these three points, so $m=3$. Next, we evaluated the definite integral $I = \int_{-2}^2 f(x) \, dx$ by first correctly defining $f(x)$ on the interval $[-2, 2]$ based on the analysis of $t^3 - 3t$. This involved splitting the integral into two parts. After performing the integration and summing the results, we obtained $I = \frac{27}{4}$. The ordered pair $(m, I)$ is $\left( 3, \frac{27}{4} \right)$.

The final answer is \boxed{\left( {3,,{{27} \over 4}} \right)}.