Question Number 194467 by horsebrand11 last updated on 08/Jul/23
$$\:\:\:\underline{\downdownarrows} \\ $$
Answered by mr W last updated on 08/Jul/23
$${f}\left({x}\right)=\sqrt{\mathrm{4}^{\mathrm{2}} −\left({x}−\mathrm{4}\right)^{\mathrm{2}} }−\sqrt{\mathrm{1}^{\mathrm{2}} −\left({x}−\mathrm{7}\right)^{\mathrm{2}} } \\ $$$${with}\:{u}={x}−\mathrm{7} \\ $$$${f}\left({u}\right)=\sqrt{\left(\mathrm{1}+\mathrm{3}\right)^{\mathrm{2}} −\left({u}+\mathrm{3}\right)^{\mathrm{2}} }−\sqrt{\mathrm{1}^{\mathrm{2}} −{u}^{\mathrm{2}} } \\ $$$${this}\:{is}\:{the}\:{difference}\:{between}\:{two}\:{circles}: \\ $$$$\left({u}+\mathrm{3}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \: \\ $$$${u}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$${f}\left({u}\right)_{{max}} \:{is}\:{at}\:{u}=−\mathrm{1} \\ $$$${f}\left({u}\right)_{{max}} =\sqrt{\mathrm{4}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{3}}\:\checkmark \\ $$$${f}\left({u}\right)_{{min}} =\mathrm{0}\:{at}\:{u}=\mathrm{1} \\ $$
Commented by mr W last updated on 08/Jul/23
Commented by horsebrand11 last updated on 08/Jul/23
$$\:\:\:\underbrace{\boldsymbol{{b}}} \\ $$
Answered by MM42 last updated on 08/Jul/23
![f(x)=(√(8−x ))( (√x)−(√(x−6)) ) ; D_f = [6 , 8] ((√(8x−x^2 ))≥(√(14x−x^2 −48)) →∀ x∈D_f ⇒ f(x)≥0 f(8)=0=min for x≥6⇒(√(8−6 ))≤ (√2) (i) −2(√(x ))(√(x−6))≤0⇒x−(x−6)−2(√(x ))(√(x−6))≤6 ⇒((√x)−(√(x−6)))^2 ≤6⇒(√x)−(√(x−6))≤(√6) (ii) (i),(ii)⇒∀ x∈D_f f(x)≤f(6)=2(√3) ⇒max_f =2(√3)](https://www.tinkutara.com/question/Q194480.png)
$${f}\left({x}\right)=\sqrt{\mathrm{8}−{x}\:}\left(\:\sqrt{{x}}−\sqrt{{x}−\mathrm{6}}\:\right)\:\:;\:\:{D}_{{f}} =\:\left[\mathrm{6}\:,\:\mathrm{8}\right] \\ $$$$\left(\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} }\geqslant\sqrt{\mathrm{14}{x}−{x}^{\mathrm{2}} −\mathrm{48}}\:\rightarrow\forall\:{x}\in{D}_{{f}} \:\Rightarrow\:{f}\left({x}\right)\geqslant\mathrm{0}\:\right. \\ $$$${f}\left(\mathrm{8}\right)=\mathrm{0}={min} \\ $$$${for}\:\:{x}\geqslant\mathrm{6}\Rightarrow\sqrt{\mathrm{8}−\mathrm{6}\:}\leqslant\:\sqrt{\mathrm{2}}\:\:\:\:\left({i}\right) \\ $$$$−\mathrm{2}\sqrt{{x}\:}\sqrt{{x}−\mathrm{6}}\leqslant\mathrm{0}\Rightarrow{x}−\left({x}−\mathrm{6}\right)−\mathrm{2}\sqrt{{x}\:}\sqrt{{x}−\mathrm{6}}\leqslant\mathrm{6} \\ $$$$\Rightarrow\left(\sqrt{{x}}−\sqrt{{x}−\mathrm{6}}\right)^{\mathrm{2}} \leqslant\mathrm{6}\Rightarrow\sqrt{{x}}−\sqrt{{x}−\mathrm{6}}\leqslant\sqrt{\mathrm{6}}\:\:\left({ii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\Rightarrow\forall\:{x}\in{D}_{{f}} \:\:\:{f}\left({x}\right)\leqslant{f}\left(\mathrm{6}\right)=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{max}_{{f}} \:=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$ \\ $$
Commented by MM42 last updated on 08/Jul/23
