Let f(x) = min {1, 1 + x sin x } for 0 leq x | vedclass

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(B) We are given $f(x) = \min \{1, 1 + x \sin x \}$.This means $f(x) = 1$ when $1 + x \sin x \geq 1$,i.e.,$x \sin x \geq 0$,and $f(x) = 1 + x \sin x$

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$(1, 0)$

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(B) We are given $f(x) = \min \{1, 1 + x \sin x \}$.This means $f(x) = 1$ when $1 + x \sin x \geq 1$,i.e.,$x \sin x \geq 0$,and $f(x) = 1 + x \sin x$ when $x \sin x In the interval $[0, 2\pi]$,$x \sin x \geq 0$ for $x \in [0, \pi]$ and $x \sin x Thus,$f(x) = 1$ for $x \in [0, \pi]$ and $f(x) = 1 + x \sin x$ for $x \in (\pi, 2\pi]$.At $x = \pi$,$f(\pi) = 1$. The left-hand limit is $\lim_{x 	o \pi^-} f(x) = 1$ and the right-hand limit is $\lim_{x 	o \pi^+} f(x) = 1 + \pi \sin(\pi) = 1$.Since the limits are equal,$f(x)$ is continuous at $x = \pi$,so $n = 0$.Now,check for differentiability at $x = \pi$:$f'(x) = 0$ for $x For $x > \pi$,$f'(x) = \sin x + x \cos x$. As $x 	o \pi^+$,$f'(x) 	o \sin(\pi) + \pi \cos(\pi) = 0 + \pi(-1) = -\pi$.Since the left-hand derivative $(0)$ is not equal to the right-hand derivative $(-\pi)$,the function is not differentiable at $x = \pi$.Thus,$m = 1$.The ordered pair $(m, n) = (1, 0)$.

Let $f(x) = \min \{1, 1 + x \sin x \}$ for $0 \leq x \leq 2\pi$. If $m$ is the number of points where $f$ is not differentiable and $n$ is the number of points where $f$ is not continuous,then the ordered pair $(m, n)$ is equal to

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