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Question-179936

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Question Number 179936 by yaslm last updated on 04/Nov/22
Answered by Frix last updated on 04/Nov/22
i f(x)= { ((x(x−2); x≤0)),((−x(x−2); 0≤x≤2)),((x(x−2); x≥2)) :} f′(x)= { ((2x−2; x≤0)),((−2x+2; 0≤x≤2)),((2x−2; x≥2)) :} f′(0)=−2∧f′(0)=2 f′(2)=−2∧f′(2)=2 both are impossible ⇒ f is not differentiable at x=0 and x=2
$${i} \\ $$$${f}\left({x}\right)=\begin{cases}{{x}\left({x}−\mathrm{2}\right);\:{x}\leqslant\mathrm{0}}\\{−{x}\left({x}−\mathrm{2}\right);\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}}\\{{x}\left({x}−\mathrm{2}\right);\:{x}\geqslant\mathrm{2}}\end{cases} \\ $$$${f}'\left({x}\right)=\begin{cases}{\mathrm{2}{x}−\mathrm{2};\:{x}\leqslant\mathrm{0}}\\{−\mathrm{2}{x}+\mathrm{2};\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}}\\{\mathrm{2}{x}−\mathrm{2};\:{x}\geqslant\mathrm{2}}\end{cases} \\ $$$${f}'\left(\mathrm{0}\right)=−\mathrm{2}\wedge{f}'\left(\mathrm{0}\right)=\mathrm{2} \\ $$$${f}'\left(\mathrm{2}\right)=−\mathrm{2}\wedge{f}'\left(\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{both}\:\mathrm{are}\:\mathrm{impossible}\:\Rightarrow \\ $$$${f}\:\mathrm{is}\:\mathrm{not}\:\mathrm{differentiable}\:\mathrm{at}\:{x}=\mathrm{0}\:\mathrm{and}\:{x}=\mathrm{2} \\ $$
Answered by Frix last updated on 04/Nov/22
iif(x)=(√(∣x∣)) = { (((√(−x)); x≤0)),(((√x); x≥0)) :} both are not differentiable at x=0 f′(x)= { ((−(1/(2(√(−x)))); x≤0 (lim at x=0 =−∞))),(((1/(2(√x))); x≥0 (lim at x=0 =+∞))) :}
$${iif}\left({x}\right)=\sqrt{\mid{x}\mid}\:=\begin{cases}{\sqrt{−{x}};\:{x}\leqslant\mathrm{0}}\\{\sqrt{{x}};\:{x}\geqslant\mathrm{0}}\end{cases} \\ $$$$\mathrm{both}\:\mathrm{are}\:\mathrm{not}\:\mathrm{differentiable}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$$${f}'\left({x}\right)=\begin{cases}{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}};\:{x}\leqslant\mathrm{0}\:\left(\mathrm{lim}\:\mathrm{at}\:{x}=\mathrm{0}\:=−\infty\right)}\\{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}};\:{x}\geqslant\mathrm{0}\:\left(\mathrm{lim}\:\mathrm{at}\:{x}=\mathrm{0}\:=+\infty\right)}\end{cases} \\ $$
Answered by Frix last updated on 04/Nov/22
iii f(x)=((sin x)/x) lim_(x→0) ((sin x)/x) =lim_(x→0) ((cos x)/1) =1 ⇒ there′s no gap in the function f′(x)=((xcos x −sin x)/x^2 ) lim_(x→0) ((xcos x −sin x)/x^2 ) =lim_(x→0) ((−xsin x)/(2x)) = =lim_(x→0) ((−sin x)/2) =0 ⇒ differentiable at x=0 ⇒ differentiable for x∈R
$${iii} \\ $$$${f}\left({x}\right)=\frac{\mathrm{sin}\:{x}}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{{x}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}}{\mathrm{1}}\:=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{gap}\:\mathrm{in}\:\mathrm{the}\:\mathrm{function} \\ $$$${f}'\left({x}\right)=\frac{{x}\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} }\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−{x}\mathrm{sin}\:{x}}{\mathrm{2}{x}}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:{x}}{\mathrm{2}}\:=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{differentiable}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{differentiable}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$
Commented by yaslm last updated on 04/Nov/22
thank you so much

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