Key Concepts and Formulas

• Continuity: A function $f(x)$ is continuous at a point $c$ if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
• Differentiability: A function $f(x)$ is differentiable at a point $c$ if the limit of the difference quotient exists, i.e., $\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$ exists. This is equivalent to the left-hand derivative being equal to the right-hand derivative at $c$: $f'(c^-) = f'(c^+)$.
• Maximum of Functions: For any real number $x$, $\max\{a, b\}$ is the larger of the two numbers $a$ and $b$. This is a piecewise function, and its points of non-differentiability often occur where the expressions being compared are equal.

Step-by-Step Solution

Step 1: Analyze the function $f(x)$ by comparing the terms $x, x^3, x^5, \ldots, x^{21}$. We need to determine which power of $x$ is the maximum for different intervals of $x$. The powers are odd integers from 1 to 21.

• For $x > 1$: $x^{21} > x^{19} > \ldots > x^3 > x$. So, $f(x) = x^{21}$.
• For $0 < x < 1$: $x > x^3 > x^5 > \ldots > x^{21}$. So, $f(x) = x$.
• For $x = 0$: $f(0) = \max\{0, 0, \ldots, 0\} = 0$.
• For $-1 < x < 0$: Let $x = -0.5$. Then $x = -0.5$, $x^3 = -0.125$, $x^5 = -0.03125$, etc. For negative odd powers, as the power increases, the absolute value decreases, making the value closer to zero. So, $x < x^3 < x^5 < \ldots < x^{21} < 0$. Thus, $f(x) = x^{21}$.
• For $x = -1$: $f(-1) = \max\{-1, (-1)^3, \ldots, (-1)^{21}\} = \max\{-1, -1, \ldots, -1\} = -1$.
• For $x < -1$: Let $x = -2$. Then $x = -2$, $x^3 = -8$, $x^5 = -32$, etc. For negative odd powers, as the power increases, the absolute value increases, making the value more negative. So, $x > x^3 > x^5 > \ldots > x^{21}$. Thus, $f(x) = x$.

Combining these observations, we can write $f(x)$ in a piecewise form:

$$f(x)=\begin{cases} x & \text{if } x < -1 \\ x^{21} & \text{if } -1 \leq x \leq 0 \\ x & \text{if } 0 < x < 1 \\ x^{21} & \text{if } x \geq 1 \end{cases}$$

Step 2: Check for continuity of $f(x)$. We need to check continuity at the points where the definition of $f(x)$ changes, which are $x = -1$, $x = 0$, and $x = 1$.

• At $x = -1$:

  • Left-hand limit: $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} x = -1$.
  • Right-hand limit: $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} x^{21} = (-1)^{21} = -1$.
  • Function value: $f(-1) = (-1)^{21} = -1$. Since $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1) = -1$, $f(x)$ is continuous at $x = -1$.

• At $x = 0$:

  • Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^{21} = 0^{21} = 0$.
  • Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$.
  • Function value: $f(0) = 0^{21} = 0$. Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$, $f(x)$ is continuous at $x = 0$.

• At $x = 1$:

  • Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$.
  • Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^{21} = 1^{21} = 1$.
  • Function value: $f(1) = 1^{21} = 1$. Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 1$, $f(x)$ is continuous at $x = 1$.

The function $f(x)$ is continuous for all $x \in \mathbb{R}$. Therefore, the number of points where $f(x)$ is not continuous, $n$, is 0.

Step 3: Find the derivative of $f(x)$ in each interval. For the intervals where $f(x)$ is defined by a single polynomial, the derivative is straightforward.

$$f'(x)=\begin{cases} \frac{d}{dx}(x) = 1 & \text{if } x < -1 \\ \frac{d}{dx}(x^{21}) = 21x^{20} & \text{if } -1 < x < 0 \\ \frac{d}{dx}(x) = 1 & \text{if } 0 < x < 1 \\ \frac{d}{dx}(x^{21}) = 21x^{20} & \text{if } x > 1 \end{cases}$$

Note that we exclude the points $-1, 0, 1$ for now, as we need to check differentiability at these points separately.

Step 4: Check for differentiability at the points where the definition of $f(x)$ changes ($x = -1$, $x = 0$, $x = 1$). For $f(x)$ to be differentiable at a point $c$, the left-hand derivative must equal the right-hand derivative at $c$.

• At $x = -1$:

  • Left-hand derivative: $f'(-1^-) = \lim_{x \to -1^-} f'(x) = \lim_{x \to -1^-} 1 = 1$.
  • Right-hand derivative: $f'(-1^+) = \lim_{x \to -1^+} f'(x) = \lim_{x \to -1^+} 21x^{20} = 21(-1)^{20} = 21(1) = 21$. Since $f'(-1^-) \neq f'(-1^+)$ ($1 \neq 21$), $f(x)$ is not differentiable at $x = -1$.

• At $x = 0$:

  • Left-hand derivative: $f'(0^-) = \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} 21x^{20} = 21(0)^{20} = 0$.
  • Right-hand derivative: $f'(0^+) = \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} 1 = 1$. Since $f'(0^-) \neq f'(0^+)$ ($0 \neq 1$), $f(x)$ is not differentiable at $x = 0$.

• At $x = 1$:

  • Left-hand derivative: $f'(1^-) = \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} 1 = 1$.
  • Right-hand derivative: $f'(1^+) = \lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} 21x^{20} = 21(1)^{20} = 21(1) = 21$. Since $f'(1^-) \neq f'(1^+)$ ($1 \neq 21$), $f(x)$ is not differentiable at $x = 1$.

The function $f(x)$ is not differentiable at $x = -1$, $x = 0$, and $x = 1$. Therefore, the number of points at which $f(x)$ is not differentiable, $m$, is 3.

Step 5: Calculate $m+n$. We found that $m = 3$ (points of non-differentiability) and $n = 0$ (points of non-continuity). So, $m+n = 3 + 0 = 3$.

Common Mistakes & Tips

• Incorrectly determining the maximum: Carefully analyze the behavior of powers of $x$ for positive, negative, and fractional values. For odd powers, the sign of $x^k$ is the same as the sign of $x$.
• Assuming differentiability if continuous: Continuity is a necessary but not sufficient condition for differentiability. Always check the equality of left-hand and right-hand derivatives.
• Forgetting to check endpoints: The points where the piecewise definition changes are critical for checking continuity and differentiability.

Summary

We first analyzed the function $f(x) = \max\{x, x^3, \ldots, x^{21}\}$ by determining which term is the maximum in different intervals of $x$. This led to a piecewise definition of $f(x)$. We then checked for continuity at the points where the definition changed ($x=-1, 0, 1$) and found that the function is continuous everywhere, so $n=0$. Next, we calculated the derivative of $f(x)$ in each interval and checked for differentiability at $x=-1, 0, 1$ by comparing the left-hand and right-hand derivatives. We found that $f(x)$ is not differentiable at these three points, so $m=3$. Finally, we calculated $m+n = 3+0=3$.

The final answer is \boxed{3}.