Let f:[0, infty) rightarrow [0, 3] be a funct | vedclass

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(A) For $0 \leq x \leq \pi$,$f(x) = \max \{\sin t : 0 \leq t \leq x\}$. Since $\sin t$ increases from $0$ to $1$ on $[0, \pi/2]$ and decreases from $1

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$f$ is differentiable everywhere in $(0, \infty)$

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(A) For $0 \leq x \leq \pi$,$f(x) = \max \{\sin t : 0 \leq t \leq x\}$. Since $\sin t$ increases from $0$ to $1$ on $[0, \pi/2]$ and decreases from $1$ to $0$ on $[\pi/2, \pi]$,we have:$f(x) = \sin x$ for $0 \leq x \leq \pi/2$,and $f(x) = 1$ for $\pi/2 For $x > \pi$,$f(x) = 2 + \cos x$.Now,check continuity at $x = \pi$:Left-hand limit: $\lim_{x 	o \pi^-} f(x) = 1$.Right-hand limit: $\lim_{x 	o \pi^+} f(x) = 2 + \cos(\pi) = 2 - 1 = 1$.Since $f(\pi) = 1$,the function is continuous at $x = \pi$.Check differentiability:At $x = \pi/2$: Left derivative is $\cos(\pi/2) = 0$,right derivative is $0$. So,$f$ is differentiable at $x = \pi/2$.At $x = \pi$: Left derivative is $0$ (since $f(x)=1$ for $x \in [\pi/2, \pi]$). Right derivative is $-\sin(\pi) = 0$. So,$f$ is differentiable at $x = \pi$.For $x > \pi$,$f(x) = 2 + \cos x$,which is differentiable everywhere. Thus,$f$ is differentiable everywhere in $(0, \infty)$.

Let $f:[0, \infty) \rightarrow [0, 3]$ be a function defined by
$f(x) = \begin{cases} \max \{\sin t : 0 \leq t \leq x\}, & 0 \leq x \leq \pi \\ 2 + \cos x, & x > \pi \end{cases}$
Then which of the following is true?

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Let $f:[0, \infty) \rightarrow [0, 3]$ be a function defined by$f(x) = \begin{cases} \max \{\sin t : 0 \leq t \leq x\}, & 0 \leq x \leq \pi \\ 2 + \cos x, & x > \pi \end{cases}$Then which of the following is true?