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Let-f-be-a-real-valued-function-defined-on-the-interval-1-1-such-that-e-x-f-x-2-0-x-t-4-1-dt-x-1-1-and-let-g-be-the-inverse-function-of-f-Find-the-value-of-g-2-




Question Number 115260 by bobhans last updated on 24/Sep/20
Let f be a real valued function defined  on the interval (−1,1) such that   e^(−x) .f(x)=2+∫_0 ^x  (√(t^4 +1)) dt ∀x∈(−1,1)  and let g be the inverse function of f  . Find the value of g′(2).
$${Let}\:{f}\:{be}\:{a}\:{real}\:{valued}\:{function}\:{defined} \\ $$$${on}\:{the}\:{interval}\:\left(−\mathrm{1},\mathrm{1}\right)\:{such}\:{that}\: \\ $$$${e}^{−{x}} .{f}\left({x}\right)=\mathrm{2}+\underset{\mathrm{0}} {\overset{{x}} {\int}}\:\sqrt{{t}^{\mathrm{4}} +\mathrm{1}}\:{dt}\:\forall{x}\in\left(−\mathrm{1},\mathrm{1}\right) \\ $$$${and}\:{let}\:{g}\:{be}\:{the}\:{inverse}\:{function}\:{of}\:{f} \\ $$$$.\:{Find}\:{the}\:{value}\:{of}\:{g}'\left(\mathrm{2}\right). \\ $$
Commented by PRITHWISH SEN 2 last updated on 24/Sep/20
Answered by john santu last updated on 24/Sep/20
Differentiating given equation we  get e^(−x) .f ′(x)−e^(−x) .f(x)=(√(1+x^4 ))  since (g○f)(x)=x as g is inverse of f  ⇒ g(f(x))=x⇒f ′(x).g ′(f(x))=1  ⇒g ′(f(0))=(1/(f ′(0)))⇒g ′(2)=(1/(f ′(0)))  ( here f(0)=2 obtained from  given equation.)  put x=0 we get f ′(0)=3.  ∴ g ′(2)=(1/3)
$${Differentiating}\:{given}\:{equation}\:{we} \\ $$$${get}\:{e}^{−{x}} .{f}\:'\left({x}\right)−{e}^{−{x}} .{f}\left({x}\right)=\sqrt{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$${since}\:\left({g}\circ{f}\right)\left({x}\right)={x}\:{as}\:{g}\:{is}\:{inverse}\:{of}\:{f} \\ $$$$\Rightarrow\:{g}\left({f}\left({x}\right)\right)={x}\Rightarrow{f}\:'\left({x}\right).{g}\:'\left({f}\left({x}\right)\right)=\mathrm{1} \\ $$$$\Rightarrow{g}\:'\left({f}\left(\mathrm{0}\right)\right)=\frac{\mathrm{1}}{{f}\:'\left(\mathrm{0}\right)}\Rightarrow{g}\:'\left(\mathrm{2}\right)=\frac{\mathrm{1}}{{f}\:'\left(\mathrm{0}\right)} \\ $$$$\left(\:{here}\:{f}\left(\mathrm{0}\right)=\mathrm{2}\:{obtained}\:{from}\right. \\ $$$$\left.{given}\:{equation}.\right) \\ $$$${put}\:{x}=\mathrm{0}\:{we}\:{get}\:{f}\:'\left(\mathrm{0}\right)=\mathrm{3}. \\ $$$$\therefore\:{g}\:'\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by mindispower last updated on 25/Sep/20
g′(f(x))=(1/(f′(x))),f′(x)=e^x (2+∫_0 ^x (√(1+t^4 )))+(√(x^4 +1))e^x )  g′(f(0))=(1/(f′(0)))=g′(2)=(1/3)
$$\left.{g}'\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{{f}'\left({x}\right)},{f}'\left({x}\right)={e}^{{x}} \left(\mathrm{2}+\int_{\mathrm{0}} ^{{x}} \sqrt{\mathrm{1}+{t}^{\mathrm{4}} }\right)+\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}{e}^{{x}} \right) \\ $$$${g}'\left({f}\left(\mathrm{0}\right)\right)=\frac{\mathrm{1}}{{f}'\left(\mathrm{0}\right)}={g}'\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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