Let f : [a, b] rightarrow [1, infty) be a con | vedclass

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(C) First,we check the continuity of $g(x)$ at $x = a$ and $x = b$.At $x = a$: $\lim_{x \rightarrow a^-} g(x) = 0$ and $\lim_{x \rightarrow a^+} g(x)

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$g(x)$ is continuous but not differentiable at $b$

Vedclass · Answer

(C) First,we check the continuity of $g(x)$ at $x = a$ and $x = b$.At $x = a$: $\lim_{x ightarrow a^-} g(x) = 0$ and $\lim_{x ightarrow a^+} g(x) = \int_a^a f(t) dt = 0$. Since $g(a) = 0$,$g(x)$ is continuous at $x = a$.At $x = b$: $\lim_{x ightarrow b^-} g(x) = \int_a^b f(t) dt$ and $\lim_{x ightarrow b^+} g(x) = \int_a^b f(t) dt$. Since $g(b) = \int_a^b f(t) dt$,$g(x)$ is continuous at $x = b$.Thus,$g(x)$ is continuous for all $x \in \mathbb{R}$.Next,we check differentiability using $g'(x) = \begin{cases} 0 & x  b \end{cases}$.At $x = a$: $g'(a^-) = 0$ and $g'(a^+) = f(a)$. Since $f(a) \in [1, \infty)$,$f(a) 
eq 0$,so $g'(a^-) 
eq g'(a^+)$. Thus,$g(x)$ is not differentiable at $x = a$.At $x = b$: $g'(b^-) = f(b)$ and $g'(b^+) = 0$. Since $f(b) \in [1, \infty)$,$f(b) 
eq 0$,so $g'(b^-) 
eq g'(b^+)$. Thus,$g(x)$ is not differentiable at $x = b$.Therefore,$g(x)$ is continuous but not differentiable at $a$ and $b$.

Let $f : [a, b] \rightarrow [1, \infty)$ be a continuous function and let $g : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $g(x) = \begin{cases} 0 & \text{if } x < a \\ \int_a^x f(t) dt & \text{if } a \leq x \leq b \\ \int_a^b f(t) dt & \text{if } x > b \end{cases}$. Then:

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