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Question Number 159642 by tounghoungko last updated on 19/Nov/21

$$minimumvalueoffunction$$$$f\left(x\right)=\sqrt{(3\sin x-4\cos x-10)(3\sin x+4\cos x-10)}$$

Commented by bobhans last updated on 19/Nov/21

$$\mathrm{g}\left(\mathrm{x}\right)=((3\sin \mathrm{x}-10)-4\cos \mathrm{x})((3\sin \mathrm{x}-10)+4\cos \mathrm{x})$$$$\mathrm{g}\left(\mathrm{x}\right)={(3\sin \mathrm{x}-10)}^2-16\cos {}^2\mathrm{x}$$$$\mathrm{g}\left(\mathrm{x}\right)=9\sin {}^2\mathrm{x}-60\sin \mathrm{x}+100-16(1-\sin {}^2\mathrm{x})$$$$\mathrm{g}\left(\mathrm{x}\right)=25\sin {}^2\mathrm{x}-60\sin \mathrm{x}+84$$$$\mathrm{g}\left(\mathrm{x}\right)={(5\sin \mathrm{x}-6)}^2+48$$$$-5⩽5\sin \mathrm{x}⩽5\Rightarrow -11⩽5\sin \mathrm{x}-6⩽-1$$$$\Rightarrow 0<{(5\sin \mathrm{x}-6)}^2⩽121$$$$\mathrm{g}{\left(\mathrm{x}\right)}_{\min }={(5.1-6)}^2+48=49$$$$\mathrm{f}{\left(\mathrm{x}\right)}_{\min }=\sqrt{49}=7$$

Answered by mr W last updated on 19/Nov/21

$$f\left(x\right)=\sqrt{{(3\sin x-10)}^2-{\left(4\cos x\right)}^2}$$$$f\left(x\right)=\sqrt{9{\sin }^{2}x+100-60\sin x-16{\cos }^{2}x}$$$$f\left(x\right)=\sqrt{25{\sin }^{2}x-60\sin x+84}$$$$f\left(x\right)=\sqrt{25{(\sin x-\frac{6}{5})}^2+48}$$$$f{\left(x\right)}_{min}=\sqrt{25{(1-\frac{6}{5})}^2+48}=7$$$$f{\left(x\right)}_{max}=\sqrt{25{(-1-\frac{6}{5})}^2+48}=13$$

Commented by mnjuly1970 last updated on 19/Nov/21

$$bravosirW$$

Answered by FongXD last updated on 19/Nov/21

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{(3\mathrm{sinx}-10-4\mathrm{cosx})(3\mathrm{sinx}-10+4\mathrm{cosx})}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{(3\mathrm{sinx}-10)}^2-{\left(4\mathrm{cosx}\right)}^2}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{9\sin }^2\mathrm{x}-60\mathrm{sinx}+100-16(1-{\sin }^2\mathrm{x})}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{25\sin }^2\mathrm{x}-60\mathrm{sinx}+84}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{(5\mathrm{sinx}-6)}^2+48}$$$$\mathrm{Min}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)\mathrm{occurs}\mathrm{when}\min \mathrm{of}\mid 5\mathrm{sinx}-6\mid \mathrm{occurs}$$$$\mathrm{we}\mathrm{have}:-1⩽\mathrm{sinx}⩽1,\mathrm{\forall }\mathrm{x}\in \mathbb{R}$$$$\iff -11⩽5\mathrm{sinx}-6⩽-1$$$$\iff 1⩽\mid 5\mathrm{sinx}-6\mid ⩽11,\Rightarrow \mathrm{Min}\mathrm{of}\mid 5\mathrm{sinx}-6\mid =1$$$$\mathrm{therefore},\mathrm{Min}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{1^2+48}=7$$

Commented by mnjuly1970 last updated on 19/Nov/21

Commented by mnjuly1970 last updated on 19/Nov/21

$$perfect$$

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