Key Concepts and Formulas

• Greatest Integer Function $[x]$: The greatest integer less than or equal to $x$. It has jump discontinuities at integer values of $x$.
• Absolute Value Function $|x-a|$: This function has a sharp corner (a point of non-differentiability) at $x=a$.
• Continuity of a function $f(x)$ at a point $c$: A function $f(x)$ is continuous at $x=c$ if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
• Differentiability of a function $f(x)$ at a point $c$: A function $f(x)$ is differentiable at $x=c$ if the left-hand derivative equals the right-hand derivative at $x=c$. That is, $\lim_{h \to 0^-} \frac{f(c+h)-f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h)-f(c)}{h}$. A function is not differentiable at a point if it is not continuous there, or if it has a sharp corner or a vertical tangent.

Step-by-Step Solution

The function is given by $f(x)=[x]+|x-2|$ for $-2 < x < 3$. We need to find the number of points of discontinuity ($m$) and non-differentiability ($n$) in the given interval.

Step 1: Analyze the components of the function. The function $f(x)$ is a sum of two functions: $[x]$ and $|x-2|$. The greatest integer function $[x]$ has discontinuities at integer values. The absolute value function $|x-2|$ has a point of non-differentiability at $x=2$.

Step 2: Determine the intervals where $[x]$ has a constant value. Within the interval $-2 < x < 3$, the integer values are $-1, 0, 1, 2$. We will break down the function definition based on the values of $[x]$ and $|x-2|$.

For $-2 < x < -1$, $[x] = -2$. For $-1 \leq x < 0$, $[x] = -1$. For $0 \leq x < 1$, $[x] = 0$. For $1 \leq x < 2$, $[x] = 1$. For $2 \leq x < 3$, $[x] = 2$.

The term $|x-2|$ can be written as: $|x-2| = -(x-2) = 2-x$ for $x < 2$. $|x-2| = x-2$ for $x \geq 2$.

Step 3: Define $f(x)$ piecewise based on the intervals.

• Interval $-2 < x < -1$: $[x] = -2$ $|x-2| = 2-x$ (since $x < 2$) $f(x) = -2 + (2-x) = -x$.

• Interval $-1 \leq x < 0$: $[x] = -1$ $|x-2| = 2-x$ (since $x < 2$) $f(x) = -1 + (2-x) = 1-x$.

• Interval $0 \leq x < 1$: $[x] = 0$ $|x-2| = 2-x$ (since $x < 2$) $f(x) = 0 + (2-x) = 2-x$.

• Interval $1 \leq x < 2$: $[x] = 1$ $|x-2| = 2-x$ (since $x < 2$) $f(x) = 1 + (2-x) = 3-x$.

• Interval $2 \leq x < 3$: $[x] = 2$ $|x-2| = x-2$ (since $x \geq 2$) $f(x) = 2 + (x-2) = x$.

So, the piecewise definition of $f(x)$ in the interval $-2 < x < 3$ is: $f(x)=\left\{\begin{array}{ll} -x, & -2< x<-1 \\ 1-x, & -1 \leq x<0 \\ 2-x, & 0 \leq x<1 \\ 3-x, & 1 \leq x<2 \\ x, & 2 \leq x<3 \end{array}\right.$

Step 4: Identify points of discontinuity. Discontinuities for $f(x)$ arise from the discontinuities of $[x]$. The integer points in the interval $-2 < x < 3$ are $x = -1, 0, 1, 2$. We need to check the continuity at these points by examining the left-hand and right-hand limits.

• At $x = -1$: $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-x) = -(-1) = 1$. $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (1-x) = 1 - (-1) = 2$. Since $\lim_{x \to -1^-} f(x) \neq \lim_{x \to -1^+} f(x)$, $f(x)$ is discontinuous at $x = -1$.

• At $x = 0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1-x) = 1 - 0 = 1$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2-x) = 2 - 0 = 2$. Since $\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$, $f(x)$ is discontinuous at $x = 0$.

• At $x = 1$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2-x) = 2 - 1 = 1$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3-x) = 3 - 1 = 2$. Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, $f(x)$ is discontinuous at $x = 1$.

• At $x = 2$: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (3-x) = 3 - 2 = 1$. $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x) = 2$. Since $\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)$, $f(x)$ is discontinuous at $x = 2$.

The function $f(x)$ is discontinuous at $x = -1, 0, 1, 2$. So, the number of points of discontinuity is $m = 4$.

Step 5: Identify points of non-differentiability. A function can be non-differentiable at points of discontinuity or at points where its derivative changes abruptly (sharp corners). From Step 4, we know that $f(x)$ is discontinuous at $x = -1, 0, 1, 2$. Therefore, it is not differentiable at these 4 points.

Now we need to check for differentiability at other points where the definition of $f(x)$ changes, which is at $x=2$ (due to the $|x-2|$ term). Let's examine the derivatives in each interval: For $-2 < x < -1$, $f(x) = -x$, so $f'(x) = -1$. For $-1 < x < 0$, $f(x) = 1-x$, so $f'(x) = -1$. For $0 < x < 1$, $f(x) = 2-x$, so $f'(x) = -1$. For $1 < x < 2$, $f(x) = 3-x$, so $f'(x) = -1$. For $2 < x < 3$, $f(x) = x$, so $f'(x) = 1$.

Now let's check differentiability at the integer points:

• At $x = -1$: Since the function is discontinuous at $x=-1$, it is not differentiable.

• At $x = 0$: Since the function is discontinuous at $x=0$, it is not differentiable.

• At $x = 1$: Since the function is discontinuous at $x=1$, it is not differentiable.

• At $x = 2$: The function is discontinuous at $x=2$, so it is not differentiable there. Let's examine the derivatives around $x=2$: Left-hand derivative at $x=2$: $\lim_{x \to 2^-} f'(x) = \lim_{x \to 2^-} (-1) = -1$. Right-hand derivative at $x=2$: $\lim_{x \to 2^+} f'(x) = \lim_{x \to 2^+} (1) = 1$. Since the left-hand and right-hand limits of the function values are not equal (from Step 4), the function is not continuous and hence not differentiable at $x=2$.

The points where $f(x)$ is not differentiable are the points of discontinuity: $x = -1, 0, 1, 2$. So, the number of points of non-differentiability is $n = 4$.

Step 6: Calculate $m+n$. We found $m=4$ (number of points of discontinuity) and $n=4$ (number of points of non-differentiability). Therefore, $m+n = 4+4 = 8$.

Common Mistakes & Tips

• Confusing discontinuity and non-differentiability: A function must be continuous at a point to be differentiable there. However, a function can be continuous but not differentiable (e.g., $|x|$ at $x=0$). In this problem, all points of non-differentiability also happen to be points of discontinuity.
• Ignoring the interval: Always ensure your piecewise function definition and analysis of continuity/differentiability are strictly within the given interval $-2 < x < 3$. The endpoints of the interval are not included, so we don't check continuity/differentiability at $x=-2$ and $x=3$.
• Incorrectly evaluating limits: Be careful when evaluating limits at integer points, especially when dealing with the greatest integer function. The limit from the left and right will often differ.

Summary The function $f(x)=[x]+|x-2|$ was analyzed by breaking it down into a piecewise function based on the integer values of $[x]$ and the point where $|x-2|$ changes definition. The points of discontinuity were identified at the integer values where $[x]$ jumps, which are $x=-1, 0, 1, 2$. Consequently, these 4 points are also points of non-differentiability. Thus, $m=4$ and $n=4$, leading to $m+n=8$.

The final answer is \boxed{8}.