Let m and n be the number of points at which | vedclass

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(A) The function is defined as $f(x) = \max \{x, x^3, x^5, \dots, x^{21}\}$.For $x \in (-1, 1)$,the maximum value is $x$ if $x > 0$ and $x^{21}$ if $x

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$3$

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(A) The function is defined as $f(x) = \max \{x, x^3, x^5, \dots, x^{21}\}$.For $x \in (-1, 1)$,the maximum value is $x$ if $x > 0$ and $x^{21}$ if $x For $|x| > 1$,the maximum value is $x$ if $x > 1$ and $x^{21}$ if $x Thus,the function can be written as:$f(x) = \begin{cases} x^{21}, & x Checking continuity:At $x = -1$: $\lim_{x 	o -1^-} f(x) = (-1)^{21} = -1$ and $\lim_{x 	o -1^+} f(x) = -1$. So,$f(x)$ is continuous at $x = -1$.At $x = 0$: $\lim_{x 	o 0^-} f(x) = 0$ and $\lim_{x 	o 0^+} f(x) = 0^{21} = 0$. So,$f(x)$ is continuous at $x = 0$.At $x = 1$: $\lim_{x 	o 1^-} f(x) = 1^{21} = 1$ and $\lim_{x 	o 1^+} f(x) = 1$. So,$f(x)$ is continuous at $x = 1$.Since $f(x)$ is continuous everywhere,$n = 0$.Checking differentiability:$f'(x) = \begin{cases} 21x^{20}, & x  1 \end{cases}$At $x = -1$: $f'(-1^-) = 21(-1)^{20} = 21$ and $f'(-1^+) = 1$. Since $21 
eq 1$,it is non-differentiable.At $x = 0$: $f'(0^-) = 1$ and $f'(0^+) = 21(0)^{20} = 0$. Since $1 
eq 0$,it is non-differentiable.At $x = 1$: $f'(1^-) = 21(1)^{20} = 21$ and $f'(1^+) = 1$. Since $21 
eq 1$,it is non-differentiable.Thus,$m = 3$.Therefore,$m + n = 3 + 0 = 3$.

Let $m$ and $n$ be the number of points at which the function $f(x) = \max \{x, x^3, x^5, \dots, x^{21}\}$,$x \in R$,is not differentiable and not continuous,respectively. Then $m + n$ is equal to . . . . . . .

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