Define g(x) = int_{-3}^3 f(x-y) f(y) , dy,for | vedclass

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(D) Given $g(x) = \int_{-3}^3 f(x-y) f(y) \, dy$ and $f(t) = \begin{cases} 1, & 0 \leq t \leq 1 \ 0, & 	ext{otherwise} \end{cases}$.Since $f(y) = 1$

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$g(x)$ is continuous everywhere and differentiable everywhere except at $x=0, 1, 2$

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(D) Given $g(x) = \int_{-3}^3 f(x-y) f(y) \, dy$ and $f(t) = \begin{cases} 1, & 0 \leq t \leq 1 \ 0, & 	ext{otherwise} \end{cases}$.Since $f(y) = 1$ only for $0 \leq y \leq 1$,the integral simplifies to $g(x) = \int_0^1 f(x-y) \, dy$.Let $t = x-y$,then $dt = -dy$. When $y=0, t=x$; when $y=1, t=x-1$.Thus,$g(x) = \int_{x-1}^x f(t) \, dt$.Evaluating this integral based on the definition of $f(t)$:If $x If $0 \leq x If $1 \leq x If $x \geq 2$,the interval $[x-1, x]$ lies outside $[0, 1]$,so $g(x) = 0$.Thus,$g(x) = \begin{cases} 0, & x \leq 0 \ x, & 0 $g(x)$ is continuous everywhere. The derivative $g'(x)$ is $\begin{cases} 0, & x  2 \end{cases}$.$g(x)$ is not differentiable at $x=0, 1, 2$ because the left and right derivatives do not match at these points.

Define $g(x) = \int_{-3}^3 f(x-y) f(y) \, dy$,for all real $x$,where $f(t) = \begin{cases} 1, & 0 \leq t \leq 1 \\ 0, & \text{otherwise} \end{cases}$. Then,

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