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Question Number 34736 by math khazana by abdo last updated on 10/May/18
letf(x)=−2x  +(√(x−3))  1)  find  f^(−1) (x)  2) calculate (f^(−1) )^′ (x)   and (f^(−1) )^, (2)  3) let g(x) = x^2  −2x+3  calculate fog(x) and give D_(fog)   4) find (fog)^(−1) (x)  5) calculate  ((fog)^(−1) )^′ (x).
$${letf}\left({x}\right)=−\mathrm{2}{x}\:\:+\sqrt{{x}−\mathrm{3}} \\ $$$$\left.\mathrm{1}\right)\:\:{find}\:\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)\:\:\:{and}\:\left({f}^{−\mathrm{1}} \right)^{,} \left(\mathrm{2}\right) \\ $$$$\left.\mathrm{3}\right)\:{let}\:{g}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:−\mathrm{2}{x}+\mathrm{3} \\ $$$${calculate}\:{fog}\left({x}\right)\:{and}\:{give}\:{D}_{{fog}} \\ $$$$\left.\mathrm{4}\right)\:{find}\:\left({fog}\right)^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{5}\right)\:{calculate}\:\:\left(\left({fog}\right)^{−\mathrm{1}} \right)^{'} \left({x}\right). \\ $$
Commented by prof Abdo imad last updated on 13/May/18
D_f =[−3,+∞[ let put y=f(x) ⇔x=f^(−1) (y)  y =f(x) ⇔ y=−2x +(√(x−3)) ⇔(y+2x)^2  =x−3  ⇒ y^2  +4yx +4x^2  −x+3=0 ⇒  4x^2   +(4y−1)x +y^2  +3 =0   Δ = (4y−1)^2  −16(y^2 +3)  = 16y^2   −8y +1 −16y^2   −48  =−8y −47  =−8(y  +((47)/8))  Δ must be≥0  ⇒ y ≤ −((47)/8)  and y +2x≥0    x_1  = ((1−4y +(√(−8y−47)))/8)  x_2 = ((1−4y −(√(−8y−47)))/8)  y +2x_1  = y   +((1−4y +(√(−8y−47)))/4)  =((1+(√(−8y−47)))/4)≥0 ⇒f^(−1) (x)= ((1−4x +(√(−8x−47)))/8)  with Df^(−1)    =]−∞,−((47)/8)]  2) (f^(−1) )^′ (x)= −(1/2) +(1/8) ((−8)/(2(√(−8x−47))))  =−(1/2) −(1/( (√(−8x−47))))     but 2 ∉ ]+∞,−((47)/8)[ so  (f^(−1) )^′ (2) dont exist  3) we have f(x)=−2x +(√(x−3 ))   and  g(x)= x^2 −2x+3 ⇒  fog(x)=−2g(x)+(√(g(x)−3))  =−2x^2  +4x −6  +(√(x^2  −2x +3−3))  =−2x^2  +4x −6  +(√(x^2  −2x))  D_(fog ) =]−∞,0]∪[2,+∞[
$${D}_{{f}} =\left[−\mathrm{3},+\infty\left[\:{let}\:{put}\:{y}={f}\left({x}\right)\:\Leftrightarrow{x}={f}^{−\mathrm{1}} \left({y}\right)\right.\right. \\ $$$${y}\:={f}\left({x}\right)\:\Leftrightarrow\:{y}=−\mathrm{2}{x}\:+\sqrt{{x}−\mathrm{3}}\:\Leftrightarrow\left({y}+\mathrm{2}{x}\right)^{\mathrm{2}} \:={x}−\mathrm{3} \\ $$$$\Rightarrow\:{y}^{\mathrm{2}} \:+\mathrm{4}{yx}\:+\mathrm{4}{x}^{\mathrm{2}} \:−{x}+\mathrm{3}=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \:\:+\left(\mathrm{4}{y}−\mathrm{1}\right){x}\:+{y}^{\mathrm{2}} \:+\mathrm{3}\:=\mathrm{0}\: \\ $$$$\Delta\:=\:\left(\mathrm{4}{y}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{16}\left({y}^{\mathrm{2}} +\mathrm{3}\right) \\ $$$$=\:\mathrm{16}{y}^{\mathrm{2}} \:\:−\mathrm{8}{y}\:+\mathrm{1}\:−\mathrm{16}{y}^{\mathrm{2}} \:\:−\mathrm{48}\:\:=−\mathrm{8}{y}\:−\mathrm{47} \\ $$$$=−\mathrm{8}\left({y}\:\:+\frac{\mathrm{47}}{\mathrm{8}}\right)\:\:\Delta\:{must}\:{be}\geqslant\mathrm{0}\:\:\Rightarrow\:{y}\:\leqslant\:−\frac{\mathrm{47}}{\mathrm{8}} \\ $$$${and}\:{y}\:+\mathrm{2}{x}\geqslant\mathrm{0}\:\:\:\:{x}_{\mathrm{1}} \:=\:\frac{\mathrm{1}−\mathrm{4}{y}\:+\sqrt{−\mathrm{8}{y}−\mathrm{47}}}{\mathrm{8}} \\ $$$${x}_{\mathrm{2}} =\:\frac{\mathrm{1}−\mathrm{4}{y}\:−\sqrt{−\mathrm{8}{y}−\mathrm{47}}}{\mathrm{8}} \\ $$$${y}\:+\mathrm{2}{x}_{\mathrm{1}} \:=\:{y}\:\:\:+\frac{\mathrm{1}−\mathrm{4}{y}\:+\sqrt{−\mathrm{8}{y}−\mathrm{47}}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}+\sqrt{−\mathrm{8}{y}−\mathrm{47}}}{\mathrm{4}}\geqslant\mathrm{0}\:\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\:\frac{\mathrm{1}−\mathrm{4}{x}\:+\sqrt{−\mathrm{8}{x}−\mathrm{47}}}{\mathrm{8}} \\ $$$$\left.{w}\left.{ith}\:{Df}^{−\mathrm{1}} \:\:\:=\right]−\infty,−\frac{\mathrm{47}}{\mathrm{8}}\right] \\ $$$$\left.\mathrm{2}\right)\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\frac{−\mathrm{8}}{\mathrm{2}\sqrt{−\mathrm{8}{x}−\mathrm{47}}} \\ $$$$\left.=−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{8}{x}−\mathrm{47}}}\:\:\:\:\:{but}\:\mathrm{2}\:\notin\:\right]+\infty,−\frac{\mathrm{47}}{\mathrm{8}}\left[\:{so}\right. \\ $$$$\left({f}^{−\mathrm{1}} \right)^{'} \left(\mathrm{2}\right)\:{dont}\:{exist} \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)=−\mathrm{2}{x}\:+\sqrt{{x}−\mathrm{3}\:}\:\:\:{and} \\ $$$${g}\left({x}\right)=\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\:\Rightarrow \\ $$$${fog}\left({x}\right)=−\mathrm{2}{g}\left({x}\right)+\sqrt{{g}\left({x}\right)−\mathrm{3}} \\ $$$$=−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:−\mathrm{6}\:\:+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:+\mathrm{3}−\mathrm{3}} \\ $$$$=−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:−\mathrm{6}\:\:+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{2}{x}} \\ $$$$\left.{D}_{{fog}\:} \left.=\right]−\infty,\mathrm{0}\right]\cup\left[\mathrm{2},+\infty\left[\right.\right. \\ $$

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