If f(x) = operatorname{Max} {3 - x, 3 + x, 6} | vedclass

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(C) The function is defined as $f(x) = \operatorname{Max} \{3 - x, 3 + x, 6\}$.To find the points of non-differentiability,we analyze the intersection

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$6$

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(C) The function is defined as $f(x) = \operatorname{Max} \{3 - x, 3 + x, 6\}$.To find the points of non-differentiability,we analyze the intersections of the functions $y_1 = 3 - x$,$y_2 = 3 + x$,and $y_3 = 6$.$1$. Intersection of $y_1$ and $y_3$: $3 - x = 6 \implies x = -3$.$2$. Intersection of $y_2$ and $y_3$: $3 + x = 6 \implies x = 3$.$3$. Intersection of $y_1$ and $y_2$: $3 - x = 3 + x \implies 2x = 0 \implies x = 0$. At $x=0$,$y_1 = 3$ and $y_2 = 3$,but $y_3 = 6$,so $f(0) = 6$.The function $f(x)$ is given by:$f(x) = \begin{cases} 3 + x, & x  3 \end{cases}$ (Wait,checking the max: for $x  6$,so $f(x) = 3-x$. For $x > 3$,$3+x > 6$,so $f(x) = 3+x$.)Correct definition: $f(x) = \begin{cases} 3 - x, & x  3 \end{cases}$.The function has sharp corners (points of non-differentiability) at $x = -3$ and $x = 3$.Thus,$a = -3$ and $b = 3$.Therefore,$|a| + |b| = |-3| + |3| = 3 + 3 = 6$.

If $f(x) = \operatorname{Max} \{3 - x, 3 + x, 6\}$ is not differentiable at $x = a$ and $x = b$,then $|a| + |b| =$

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