Question Number 59188 by maxmathsup by imad last updated on 05/May/19
$${let}\:{f}\left({x}\right)={x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:\:{and}\:{g}\left({x}\right)\:=\frac{\mathrm{2}\:+\sqrt{{x}−\mathrm{3}}}{\mathrm{2}−\sqrt{{x}−\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:\:{D}_{{f}} \:\:,{D}_{{g}} \:\:\:{and}\:{D}_{{fog}} \:\:\:\:\:{and}\:\:{determine}\:{fog}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{gof}\left({x}\right)\:{and}\:{give}\:{D}_{{gof}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right){dx}\:\:\:\: \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\:\int_{\mathrm{4}} ^{\mathrm{5}} \:{g}\left({x}\right){dx}\:. \\ $$
Commented by maxmathsup by imad last updated on 06/May/19
![1) x∈D_f ⇔4−x^2 ≥0 ⇔x^2 ≤4 ⇔∣x∣≤2 ⇔−2≤x≤2 ⇒D_f =[−2,2] x∈D_g ⇔x−3≥0 and 2−(√(x−3))≠0 ⇔x≥3 and (√(x−3))≠2 ⇔x≥3 and x≠7 ⇒ D_g =[3,7[∪]7,+∞ [ x∈fog ⇔ x∈D_g and g(x)∈ D_f ⇔ x≥3 and x≠7 and −2≤g(x)≤2 but −2≤g(x)≤2 ⇒ 4−g^2 (x)≥0 ⇒4−(((2+(√(x−3)))/(2−(√(x−3)))))^2 ≥0 ⇒ ((4(2−(√(x−3)))^2 −(2+(√(x−3)))^2 )/((2−(√(x−3)))^2 )) ≥0 ⇒4(4−4(√(x−3))+x−3)−(4+4(√(x−3))+x−3)≥0 ⇒16−16(√(x−3))+4x−12 −4−4(√(x−3))−x+3 ≥0 ⇒ −20(√(x−3)) +3x +3 ≥0 ⇒3x+3 ≥20(√(x−3)) ⇒(3x+3)^2 ≥400(x−3) ⇒ 9(x^2 +2x+1)≥400x−1200 ⇒ 9x^2 +18x−400x +9 +1200 ≥0 ⇒9x^2 −382x +1209 ≥0 Δ^′ =(191)^2 −9.1209 =....](https://www.tinkutara.com/question/Q59229.png)
$$\left.\mathrm{1}\right)\:{x}\in{D}_{{f}} \Leftrightarrow\mathrm{4}−{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\Leftrightarrow{x}^{\mathrm{2}} \leqslant\mathrm{4}\:\Leftrightarrow\mid{x}\mid\leqslant\mathrm{2}\:\Leftrightarrow−\mathrm{2}\leqslant{x}\leqslant\mathrm{2}\:\Rightarrow{D}_{{f}} =\left[−\mathrm{2},\mathrm{2}\right] \\ $$$${x}\in{D}_{{g}} \:\Leftrightarrow{x}−\mathrm{3}\geqslant\mathrm{0}\:{and}\:\mathrm{2}−\sqrt{{x}−\mathrm{3}}\neq\mathrm{0}\:\Leftrightarrow{x}\geqslant\mathrm{3}\:{and}\:\sqrt{{x}−\mathrm{3}}\neq\mathrm{2}\:\Leftrightarrow{x}\geqslant\mathrm{3}\:{and}\:{x}\neq\mathrm{7}\:\Rightarrow \\ $$$${D}_{{g}} =\left[\mathrm{3},\mathrm{7}\left[\cup\right]\mathrm{7},+\infty\:\left[\right.\right. \\ $$$${x}\in{fog}\:\Leftrightarrow\:{x}\in{D}_{{g}} {and}\:{g}\left({x}\right)\in\:{D}_{{f}} \:\Leftrightarrow\:{x}\geqslant\mathrm{3}\:{and}\:{x}\neq\mathrm{7}\:\:{and}\:\:\:−\mathrm{2}\leqslant{g}\left({x}\right)\leqslant\mathrm{2}\:\:{but} \\ $$$$−\mathrm{2}\leqslant{g}\left({x}\right)\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{4}−{g}^{\mathrm{2}} \left({x}\right)\geqslant\mathrm{0}\:\Rightarrow\mathrm{4}−\left(\frac{\mathrm{2}+\sqrt{{x}−\mathrm{3}}}{\mathrm{2}−\sqrt{{x}−\mathrm{3}}}\right)^{\mathrm{2}} \:\geqslant\mathrm{0}\:\Rightarrow \\ $$$$\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{2}} −\left(\mathrm{2}+\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{2}} }{\left(\mathrm{2}−\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{2}} }\:\geqslant\mathrm{0}\:\Rightarrow\mathrm{4}\left(\mathrm{4}−\mathrm{4}\sqrt{{x}−\mathrm{3}}+{x}−\mathrm{3}\right)−\left(\mathrm{4}+\mathrm{4}\sqrt{{x}−\mathrm{3}}+{x}−\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{16}−\mathrm{16}\sqrt{{x}−\mathrm{3}}+\mathrm{4}{x}−\mathrm{12}\:−\mathrm{4}−\mathrm{4}\sqrt{{x}−\mathrm{3}}−{x}+\mathrm{3}\:\geqslant\mathrm{0}\:\Rightarrow \\ $$$$−\mathrm{20}\sqrt{{x}−\mathrm{3}}\:+\mathrm{3}{x}\:+\mathrm{3}\:\geqslant\mathrm{0}\:\Rightarrow\mathrm{3}{x}+\mathrm{3}\:\geqslant\mathrm{20}\sqrt{{x}−\mathrm{3}}\:\Rightarrow\left(\mathrm{3}{x}+\mathrm{3}\right)^{\mathrm{2}} \geqslant\mathrm{400}\left({x}−\mathrm{3}\right)\:\Rightarrow \\ $$$$\mathrm{9}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)\geqslant\mathrm{400}{x}−\mathrm{1200}\:\Rightarrow \\ $$$$\mathrm{9}{x}^{\mathrm{2}} +\mathrm{18}{x}−\mathrm{400}{x}\:+\mathrm{9}\:+\mathrm{1200}\:\geqslant\mathrm{0}\:\Rightarrow\mathrm{9}{x}^{\mathrm{2}} \:−\mathrm{382}{x}\:+\mathrm{1209}\:\geqslant\mathrm{0} \\ $$$$\Delta^{'} \:=\left(\mathrm{191}\right)^{\mathrm{2}} −\mathrm{9}.\mathrm{1209}\:=…. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 06/May/19
![fog(x) =f(g(x)) =g(x)−(√(4−(g(x))^2 ))=((2+(√(x−3)))/(2−(√(x−3)))) −(√(4−(((2+(√(x−3)))/(2−(√(x−3)))))^2 )) 2) gof(x) =g(f(x)) =((2+(√(f(x)−3)))/(2−(√(f(x)−3)))) =((2−(√(x−(√(4−x^2 ))−3)))/(2−(√(x−(√(4−x^2 ))−3)))) x∈D_(g0f) ⇔ x−3−(√(4−x^2 ))≥0 and 4−x^2 ≥0 and (√(x−3−(√(4−x^2 ))))≠2 ⇒ but 4−x^2 ≥0 ⇒−2≤x≤2 x−3−(√(4−x^2 ))≥0 ⇒x−3≥(√(4−x^2 )) we must have x≥3 but x∈[−2,2] condition impossible so gof is not defined ...!](https://www.tinkutara.com/question/Q59231.png)
$${fog}\left({x}\right)\:={f}\left({g}\left({x}\right)\right)\:={g}\left({x}\right)−\sqrt{\mathrm{4}−\left({g}\left({x}\right)\right)^{\mathrm{2}} }=\frac{\mathrm{2}+\sqrt{{x}−\mathrm{3}}}{\mathrm{2}−\sqrt{{x}−\mathrm{3}}}\:−\sqrt{\mathrm{4}−\left(\frac{\mathrm{2}+\sqrt{{x}−\mathrm{3}}}{\mathrm{2}−\sqrt{{x}−\mathrm{3}}}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{gof}\left({x}\right)\:={g}\left({f}\left({x}\right)\right)\:=\frac{\mathrm{2}+\sqrt{{f}\left({x}\right)−\mathrm{3}}}{\mathrm{2}−\sqrt{{f}\left({x}\right)−\mathrm{3}}}\:=\frac{\mathrm{2}−\sqrt{{x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }−\mathrm{3}}}{\mathrm{2}−\sqrt{{x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }−\mathrm{3}}} \\ $$$${x}\in{D}_{{g}\mathrm{0}{f}} \:\Leftrightarrow\:{x}−\mathrm{3}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\geqslant\mathrm{0}\:{and}\:\mathrm{4}−{x}^{\mathrm{2}} \geqslant\mathrm{0}\:{and}\:\sqrt{{x}−\mathrm{3}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\neq\mathrm{2}\:\Rightarrow \\ $$$${but}\:\:\mathrm{4}−{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow−\mathrm{2}\leqslant{x}\leqslant\mathrm{2} \\ $$$${x}−\mathrm{3}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\geqslant\mathrm{0}\:\Rightarrow{x}−\mathrm{3}\geqslant\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:\:\:{we}\:{must}\:{have}\:{x}\geqslant\mathrm{3}\:{but}\:\:{x}\in\left[−\mathrm{2},\mathrm{2}\right] \\ $$$${condition}\:{impossible}\:\:{so}\:{gof}\:{is}\:{not}\:{defined}\:…! \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 06/May/19
![3) ∫_(−(1/2)) ^(1/2) f(x)dx =∫_(−(1/2)) ^(1/2) (x−(√(4−x^2 )))dx =∫_(−(1/2)) ^(1/2) xdx −2∫_0 ^(1/2) (√(4−x^2 ))dx =0−2∫_0 ^(1/2) (√(4−x^2 ))dx =_(x=2sint) −2 ∫_0 ^(arcsin((1/4))) 2(√(1−sin^2 t))(2cost)dt =−8 ∫_0 ^(arcsin((1/4))) cos^2 t dt =−4 ∫_0 ^(arcsin((1/4))) (1+cos(2t))dt =−4 arcsin((1/4)) −2 [sin(2t)]_0 ^(arcsin((1/4))) =−4 arcsin((1/4))−2{ sin(2arsin((1/4))} but sin(2arcsin((1/4)))=2sin(arcsin((1/4)))(√(1−sin^2 (arsin((1/4))))) =(1/2)(√(1−((1/4))^2 ))=(1/2)(√(1−(1/(16))))=(1/2) ((√(15))/4) =((√(15))/8) ⇒ ∫_(−(1/2)) ^(1/2) f(x)dx =−4 arcsin((1/4))−((√(15))/4)](https://www.tinkutara.com/question/Q59232.png)
$$\left.\mathrm{3}\right)\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right){dx}\:=\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\right){dx}\:=\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:{xdx}\:−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{0}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:=_{{x}=\mathrm{2}{sint}} \:\:\:\:−\mathrm{2}\:\int_{\mathrm{0}} ^{{arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \mathrm{2}\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}\left(\mathrm{2}{cost}\right){dt} \\ $$$$=−\mathrm{8}\:\int_{\mathrm{0}} ^{{arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} {cos}^{\mathrm{2}} {t}\:{dt}\:=−\mathrm{4}\:\int_{\mathrm{0}} ^{{arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \:\left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=−\mathrm{4}\:{arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:−\mathrm{2}\:\left[{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$=−\mathrm{4}\:{arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{2}\left\{\:{sin}\left(\mathrm{2}{arsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\}\:\:{but}\:\right. \\ $$$${sin}\left(\mathrm{2}{arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)=\mathrm{2}{sin}\left({arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({arsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\:=\frac{\sqrt{\mathrm{15}}}{\mathrm{8}}\:\Rightarrow \\ $$$$\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right){dx}\:=−\mathrm{4}\:{arcsin}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 06/May/19
![4) ∫_4 ^5 g(x)dx =∫_4 ^5 ((2+(√(x−3)))/(2−(√(x−3)))) dx =_((√(x−3))=t) ∫_1 ^(√2) ((2+t)/(2−t))(2t)dt =2 ∫_1 ^(√2) ((t^2 +2t)/(2−t)) dt =−2 ∫_1 ^(√2) ((t^2 +2t)/(t−2)) dt =−2∫_1 ^(√2) ((t^2 −4 +4+2t)/(t−2)) =−2∫_1 ^(√2) ( t+2 +((2t+4)/(t−2)))dt =−2 ∫_1 ^(√2) (t+2)dt −4 ∫_1 ^(√2) ((t+2)/(t−2)) dt but ∫_1 ^(√2) (t+2)dt =[(t^2 /2) +2t]_1 ^(√2) =1+2(√2)−(1/2) −2 =2(√2)−(3/2) ∫_1 ^(√2) ((t+2)/(t−2)) dt =∫_1 ^(√2) ((t−2+4)/(t−2)) dt =(√2)−1 +4[ln∣t−2∣]_1 ^(√2) =(√2)−1 +4(ln(2−(√2)) ⇒ ∫_4 ^5 g(x)dx =−4(√2) +3 −4(√2) +4−16ln(2−(√2))=−8(√2) +7−16ln(2−(√2)) .](https://www.tinkutara.com/question/Q59233.png)
$$\left.\mathrm{4}\right)\:\int_{\mathrm{4}} ^{\mathrm{5}} {g}\left({x}\right){dx}\:=\int_{\mathrm{4}} ^{\mathrm{5}} \:\frac{\mathrm{2}+\sqrt{{x}−\mathrm{3}}}{\mathrm{2}−\sqrt{{x}−\mathrm{3}}}\:{dx}\:=_{\sqrt{{x}−\mathrm{3}}={t}} \:\:\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{\mathrm{2}+{t}}{\mathrm{2}−{t}}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}}{\mathrm{2}−{t}}\:{dt}\:=−\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{2}{t}}{{t}−\mathrm{2}}\:{dt}\:=−\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{t}^{\mathrm{2}} −\mathrm{4}\:+\mathrm{4}+\mathrm{2}{t}}{{t}−\mathrm{2}} \\ $$$$=−\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left(\:\:{t}+\mathrm{2}\:\:+\frac{\mathrm{2}{t}+\mathrm{4}}{{t}−\mathrm{2}}\right){dt}\:=−\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left({t}+\mathrm{2}\right){dt}\:\:−\mathrm{4}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{t}+\mathrm{2}}{{t}−\mathrm{2}}\:{dt}\:{but} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left({t}+\mathrm{2}\right){dt}\:=\left[\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{2}{t}\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:=\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:−\mathrm{2}\:=\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{{t}+\mathrm{2}}{{t}−\mathrm{2}}\:{dt}\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{t}−\mathrm{2}+\mathrm{4}}{{t}−\mathrm{2}}\:{dt}\:=\sqrt{\mathrm{2}}−\mathrm{1}\:+\mathrm{4}\left[{ln}\mid{t}−\mathrm{2}\mid\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} =\sqrt{\mathrm{2}}−\mathrm{1}\:+\mathrm{4}\left({ln}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\:\Rightarrow\right. \\ $$$$\int_{\mathrm{4}} ^{\mathrm{5}} {g}\left({x}\right){dx}\:=−\mathrm{4}\sqrt{\mathrm{2}}\:+\mathrm{3}\:−\mathrm{4}\sqrt{\mathrm{2}}\:+\mathrm{4}−\mathrm{16}{ln}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)=−\mathrm{8}\sqrt{\mathrm{2}}\:+\mathrm{7}−\mathrm{16}{ln}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\:. \\ $$
Answered by Forkum Michael Choungong last updated on 05/May/19
$$\left.{for}\:\mathrm{1}\right) \\ $$$$\left.\mathrm{1}\right)\:{f}\left({x}\right)=\:{x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{and}\:{g}\left({x}\right)\:=\frac{\mathrm{2}+\sqrt{{x}−\mathrm{3}}}{\mathrm{2}−\sqrt{{x}−\mathrm{3}}} \\ $$$${let}\:{f}\left({x}\right)=\mathrm{0} \\ $$$${x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }={x} \\ $$$$\mathrm{4}−{x}^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{4}=\mathrm{2}{x}^{\mathrm{2}} \\ $$$${x}=\sqrt{\mathrm{2}} \\ $$$${D}_{{f}\:} =\:\left\{\:{x}:{x}\:\in\:\mathbb{R}\:,\:{x}\neq\sqrt{\mathrm{2}}\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Mr X pcx last updated on 06/May/19
$${not}\:{correct}\:{sir}… \\ $$
Answered by MJS last updated on 06/May/19
$${D}_{{f}} =\left\{{x}\in\mathbb{R}\:\mid\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{2}\right\};\:{R}_{{f}} =\left\{{y}\in\mathbb{R}\:\mid\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{y}\leqslant\mathrm{2}\right\} \\ $$$${D}_{{g}} =\left\{{x}\in\mathbb{R}\:\mid\:{x}\geqslant\mathrm{3}\wedge{x}\neq\mathrm{7}\right\};\:{R}_{{g}} =\left\{{y}\in\mathbb{R}\:\mid\:\mid{y}\mid\geqslant\mathrm{1}\right\} \\ $$$${f}\left({g}\left({x}\right)\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:−\mathrm{2}\leqslant{g}\left({x}\right)\leqslant\mathrm{2}\:\Rightarrow \\ $$$$\Rightarrow\:{D}_{{f}\left({g}\left({x}\right)\right)} =\left\{{x}\in\mathbb{R}\:\mid\:\mathrm{3}\leqslant{x}\leqslant\frac{\mathrm{31}}{\mathrm{9}}\:\vee\:{x}\geqslant\mathrm{39}\right\} \\ $$$${g}\left({f}\left({x}\right)\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{f}\left({x}\right)\geqslant\mathrm{3}\wedge{f}\left({x}\right)\neq\mathrm{7}\:\Rightarrow \\ $$$$\Rightarrow\:{D}_{{g}\left({f}\left({x}\right)\right)} =\left\{\right\} \\ $$