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Question Number 125995 by liberty last updated on 16/Dec/20

Commented by benjo_mathlover last updated on 16/Dec/20

f(x)=  { ((x ; 0≤x≤1)),((−x ; −1≤x≤0)) :}   f ′(−1) = lim_(h→0)  ((f(−1+h)−f(−1))/h)  f ′(−1)= lim_(h→0) ((−(−1+h)−(1))/h)   f ′(−1)= lim_(h→0)  ((−h)/h) = −1

$${f}\left({x}\right)=\:\begin{cases}{{x}\:;\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}}\\{−{x}\:;\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{0}}\end{cases} \\ $$$$\:{f}\:'\left(−\mathrm{1}\right)\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left(−\mathrm{1}+{h}\right)−{f}\left(−\mathrm{1}\right)}{{h}} \\ $$$${f}\:'\left(−\mathrm{1}\right)=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\left(−\mathrm{1}+{h}\right)−\left(\mathrm{1}\right)}{{h}} \\ $$$$\:{f}\:'\left(−\mathrm{1}\right)=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−{h}}{{h}}\:=\:−\mathrm{1} \\ $$

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