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Suppose $f(x)$ is differentiable on $U=(-\infty, 0]$ and $g(x)$ is not differentiable at $V=[0,\infty)$ but is differentiable on $(0,\infty)$. Is the following piecewise function $h(x)$ differentiable? $$ h(x)= \begin{cases} f(x) \text{ for } x\in U\\ g(x) \text{ for } x\in V\\ \end{cases} $$ Note $U\cap V=\{0\}$ and $f$ is differentiable at $0$ but $g$ is not. Disclaimer, I'm not sure why you would want to consider a piecewise function with "overlapping pieces" but I was asked to.

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  • $\begingroup$ What have you tried so far? What do you think the answer is? Does your intuition tell you that $h$ is differentiable or not? How would you go about proving what your intuition tells you? $\endgroup$ – James Sep 22 '16 at 0:48
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Hint: For $t > 0$, $$ \frac{h(0+t) - h(0)}{t} = \frac{g(0+t) - g(0)}{t} $$

Also, it is common to construct a "global" function by piecing together smaller (often more "local") functions. So, this construction is not solely for the point of pedagogy.

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The condition on $g$ tells you that $g$ is not differentiable at the point $0$.

Then examine derivative from the right $$ \lim_{x \rightarrow 0^+}\frac{g(x)-g(0)}{x} $$ Which you already know does not exist.

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