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Question Number 40102 by maxmathsup by imad last updated on 15/Jul/18

let f(x) = ((∣x∣)/((1+∣1−x^2 ∣)^n )) study tbe derivability of f at points 0 and 1 (n natural integr)

$${let}\:{f}\left({x}\right)\:=\:\frac{\mid{x}\mid}{\left(\mathrm{1}+\mid\mathrm{1}−{x}^{\mathrm{2}} \mid\right)^{{n}} } \\ $$$${study}\:{tbe}\:{derivability}\:{of}\:{f}\:{at}\:{points}\:\mathrm{0}\:{and}\:\mathrm{1}\:\left({n}\:{natural}\:{integr}\right) \\ $$

Answered by math khazana by abdo last updated on 26/Jul/18

1) derivability at 0 lim_(h→0^+ ) ((f(0+h) −f(0))/h) =lim_(h→0^+ ) (h/((1+(1−h^2 )^n )) =lim_(h→0^+ ) (h/(h(2+h^2 )^n )) =lim_(h→0^+ ) (1/((2+h^2 )^n )) =(1/2^n ) lim_(h→0^− ) ((f(0+h) −f(0))/h) =lim_(h→0^− ) ((−1)/((2+h^2 )^n )) =((−1)/2^n ) we conclude that f is derivable at left and right at but not derivable at this point.

$$\left.\mathrm{1}\right)\:{derivability}\:{at}\:\mathrm{0} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{{f}\left(\mathrm{0}+{h}\right)\:−{f}\left(\mathrm{0}\right)}{{h}}\:={lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{{h}}{\left(\mathrm{1}+\left(\mathrm{1}−{h}^{\mathrm{2}} \right)^{{n}} \right.} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\:\:\:\:\:\frac{{h}}{{h}\left(\mathrm{2}+{h}^{\mathrm{2}} \right)^{{n}} }\:={lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}+{h}^{\mathrm{2}} \right)^{{n}} }\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$${lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\frac{{f}\left(\mathrm{0}+{h}\right)\:−{f}\left(\mathrm{0}\right)}{{h}}\:={lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\:\frac{−\mathrm{1}}{\left(\mathrm{2}+{h}^{\mathrm{2}} \right)^{{n}} }\:=\frac{−\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$${we}\:{conclude}\:{that}\:{f}\:{is}\:{derivable}\:{at}\:{left}\:{and} \\ $$$${right}\:{at}\:{but}\:{not}\:{derivable}\:{at}\:{this}\:{point}. \\ $$$$ \\ $$$$\: \\ $$

Answered by math khazana by abdo last updated on 28/Jul/18

2) derivability at 1 lim_(h→0^+ ) ((f(1+h) −f(1))/h) =lim_(h→0^+ ) ((∣1+h∣)/((1+∣1−(1+h)^2 ∣)^n )) −1 =lim_(h→0^+ ) ((1+h)/((1+((1+h)^2 −1))^n )) −1 =lim_(h→0^+ ) ((1+h)/((1+h^2 +2h)^n )) −1 =lim_(h→0^− ) ((1+h−(h+1)^(2n) )/((h+1)^(2n) )) =0 f is derivable at right of 1 lim_(h→0^− ) ((f(1+h)−f(1))/h)=lim_(h→0^− ) ((1+h)/({1+∣1−(1+h)^2 }^n )) −1 =lim_(h→0^− ) ((1+h)/({1+(1−(1+h)^2 }^n )) −1 =lim_(h→0^− ) ((1+h)/({1+(1−h^2 −2h−1)}^n )) −1 =lim_(h→0^− ) ((1+h)/({1−h^2 −2h)^n ))−1 =0 f is derivable at left of 1 we have f_g ^′ (1)=f_d ^′ (1) ⇒ f is derivable at x_0 =1

$$\left.\mathrm{2}\right)\:{derivability}\:{at}\:\mathrm{1} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\frac{{f}\left(\mathrm{1}+{h}\right)\:−{f}\left(\mathrm{1}\right)}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\frac{\mid\mathrm{1}+{h}\mid}{\left(\mathrm{1}+\mid\mathrm{1}−\left(\mathrm{1}+{h}\right)^{\mathrm{2}} \mid\right)^{{n}} }\:−\mathrm{1} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{\mathrm{1}+{h}}{\left(\mathrm{1}+\left(\left(\mathrm{1}+{h}\right)^{\mathrm{2}} −\mathrm{1}\right)\right)^{{n}} }\:−\mathrm{1} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{\mathrm{1}+{h}}{\left(\mathrm{1}+{h}^{\mathrm{2}} \:+\mathrm{2}{h}\right)^{{n}} }\:−\mathrm{1} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\:\frac{\mathrm{1}+{h}−\left({h}+\mathrm{1}\right)^{\mathrm{2}{n}} }{\left({h}+\mathrm{1}\right)^{\mathrm{2}{n}} }\:=\mathrm{0}\:\:{f}\:{is}\:{derivable}\:{at} \\ $$$${right}\:{of}\:\mathrm{1} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\:\frac{{f}\left(\mathrm{1}+{h}\right)−{f}\left(\mathrm{1}\right)}{{h}}={lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\:\frac{\mathrm{1}+{h}}{\left\{\mathrm{1}+\mid\mathrm{1}−\left(\mathrm{1}+{h}\right)^{\mathrm{2}} \right\}^{{n}} }\:−\mathrm{1} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\:\:\:\frac{\mathrm{1}+{h}}{\left\{\mathrm{1}+\left(\mathrm{1}−\left(\mathrm{1}+{h}\right)^{\mathrm{2}} \right\}^{{n}} \right.}\:−\mathrm{1} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\:\:\frac{\mathrm{1}+{h}}{\left\{\mathrm{1}+\left(\mathrm{1}−{h}^{\mathrm{2}} −\mathrm{2}{h}−\mathrm{1}\right)\right\}^{{n}} }\:−\mathrm{1} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\:\:\frac{\mathrm{1}+{h}}{\left\{\mathrm{1}−{h}^{\mathrm{2}} −\mathrm{2}{h}\right)^{{n}} }−\mathrm{1}\:=\mathrm{0} \\ $$$${f}\:{is}\:{derivable}\:{at}\:{left}\:{of}\:\mathrm{1}\:\:{we}\:{have} \\ $$$${f}_{{g}} ^{'} \left(\mathrm{1}\right)={f}_{{d}} ^{'} \left(\mathrm{1}\right)\:\Rightarrow\:{f}\:{is}\:{derivable}\:{at}\:{x}_{\mathrm{0}} =\mathrm{1} \\ $$$$ \\ $$

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