The function f(x) = max {(1 - x), (1 + x), 2} | vedclass

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(C) Given $f(x) = \max \{(1 - x), (1 + x), 2\}.$We can define $f(x)$ piecewise by analyzing the intervals:If $x > 1$,then $1 + x > 2$ and $1 + x > 1 -

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Differentiable at all points except at $x = 1$ and $x = - 1$

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(C) Given $f(x) = \max \{(1 - x), (1 + x), 2\}.$We can define $f(x)$ piecewise by analyzing the intervals:If $x > 1$,then $1 + x > 2$ and $1 + x > 1 - x$,so $f(x) = 1 + x.$If $- 1 \le x \le 1$,then $2 \ge 1 + x$ and $2 \ge 1 - x$,so $f(x) = 2.$If $x  2$ and $1 - x > 1 + x$,so $f(x) = 1 - x.$Thus,$f(x) = \begin{cases} 1 - x, & x  1 \end{cases}$Since $f(x)$ is a polynomial function in each interval and the pieces match at the boundaries ($f(-1) = 2$ and $f(1) = 2$),the function is continuous everywhere.Checking differentiability at $x = - 1$:Left-hand derivative: $\frac{d}{dx}(1 - x) = - 1.$Right-hand derivative: $\frac{d}{dx}(2) = 0.$Since $- 1 
eq 0$,it is not differentiable at $x = - 1.$Checking differentiability at $x = 1$:Left-hand derivative: $\frac{d}{dx}(2) = 0.$Right-hand derivative: $\frac{d}{dx}(1 + x) = 1.$Since $0 
eq 1$,it is not differentiable at $x = 1.$Therefore,the function is differentiable at all points except at $x = 1$ and $x = - 1$.

The function $f(x) = \max \{(1 - x), (1 + x), 2\},$ $x \in ( - \infty , \infty ),$ is

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