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I was just wondering what the derivative of $f(x) = \max(0,1-x)^{2}$ would be. What technique do you use to determine this derivative?

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Don't you mean $\max(0, (1-x)^2)$? – Javier Apr 21 '13 at 19:05
since $(1-x)^2 \ge 0 $ (for x a real number), $\max(0,(1-x)^2)$ is equal to $(1-x)^2$ – Andre Holzner Jun 8 at 12:35
up vote 16 down vote accepted

It might be of help to sketch the function or write it without the $\max$. We get $$f(x) = \begin{cases} (1-x)^2 & \text{if } x \leq 1\\ 0 & \text{if }x \geq 1\end{cases}$$ It is easy to work out the derivative everywhere except at $x=1$. At $x=1$, work out explicitly from definition. $$\lim_{h \to 0^+} \dfrac{f(1+h) - f(1)}{h} = 0$$ $$\lim_{h \to 0^-} \dfrac{f(1+h) - f(1)}{h} = \lim_{h\to 0^-} \dfrac{h^2}{h} = 0$$ Hence, we have $$f'(x) = \begin{cases} 2(x-1) & \text{if } x \leq 1\\ 0 & \text{if }x \geq 1\end{cases}$$

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HINT : Analyze this function on different intervals.

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If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)? – phil12 Apr 21 '13 at 18:31
Now what happens when x = 1 ? Does f'(1) exist ? – Kalissar Apr 21 '13 at 18:33

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