If both f(x) and g(x) are differentiable func | vedclass

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(C) The function $h(x) = \max\{f(x), g(x)\}$ can be written as $h(x) = \frac{f(x) + g(x) + |f(x) - g(x)|}{2}$.Since $f(x)$ and $g(x)$ are differentiab

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is differentiable at $x = x_0$ provided $f(x_0) 
eq g(x_0)$

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(C) The function $h(x) = \max\{f(x), g(x)\}$ can be written as $h(x) = \frac{f(x) + g(x) + |f(x) - g(x)|}{2}$.Since $f(x)$ and $g(x)$ are differentiable at $x = x_0$,their sum and difference are also differentiable.The differentiability of $h(x)$ depends on the term $|f(x) - g(x)|$.If $f(x_0) 
eq g(x_0)$,then by the continuity of $f$ and $g$,there exists a neighborhood around $x_0$ where $f(x) - g(x)$ maintains a constant sign,making $|f(x) - g(x)|$ differentiable at $x_0$.If $f(x_0) = g(x_0)$,the function $h(x)$ is differentiable at $x_0$ if and only if $f'(x_0) = g'(x_0)$.Therefore,$h(x)$ is guaranteed to be differentiable at $x_0$ if $f(x_0) 
eq g(x_0)$.

If both $f(x)$ and $g(x)$ are differentiable functions at $x = x_0$,then the function defined as $h(x) = \text{Maximum} \{f(x), g(x)\}$:

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