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Question Number 194467 by horsebrand11 last updated on 08/Jul/23

$$\underset{―}{\downdownarrows }$$

Answered by mr W last updated on 08/Jul/23

$$f\left(x\right)=\sqrt{4^2-{(x-4)}^2}-\sqrt{1^2-{(x-7)}^2}$$$$withu=x-7$$$$f\left(u\right)=\sqrt{{(1+3)}^2-{(u+3)}^2}-\sqrt{1^2-u^2}$$$$thisisthedifferencebetweentwocircles:$$$${(u+3)}^2+y^2=4^2$$$$u^2+y^2=1^2$$$$f{\left(u\right)}_{max}isatu=-1$$$$f{\left(u\right)}_{max}=\sqrt{4^2-2^2}=2\sqrt{3}✓$$$$f{\left(u\right)}_{min}=0atu=1$$

Commented by mr W last updated on 08/Jul/23

Commented by horsebrand11 last updated on 08/Jul/23

$$\underbrace{\mathit{b}}$$

Answered by MM42 last updated on 08/Jul/23

$$f\left(x\right)=\sqrt{8-x}(\sqrt{x}-\sqrt{x-6});D_f=[6,8]$$$$(\sqrt{8x-x^2}⩾\sqrt{14x-x^2-48}\to \mathrm{\forall }x\in D_f\Rightarrow f\left(x\right)⩾0$$$$f\left(8\right)=0=min$$$$forx⩾6\Rightarrow \sqrt{8-6}⩽\sqrt{2}\left(i\right)$$$$-2\sqrt{x}\sqrt{x-6}⩽0\Rightarrow x-(x-6)-2\sqrt{x}\sqrt{x-6}⩽6$$$$\Rightarrow {(\sqrt{x}-\sqrt{x-6})}^2⩽6\Rightarrow \sqrt{x}-\sqrt{x-6}⩽\sqrt{6}\left(ii\right)$$$$\left(i\right),\left(ii\right)\Rightarrow \mathrm{\forall }x\in D_{f}f\left(x\right)⩽f\left(6\right)=2\sqrt{3}$$$$\Rightarrow {max}_f=2\sqrt{3}$$$$$$

Commented by MM42 last updated on 08/Jul/23

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