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Question Number 159642 by tounghoungko last updated on 19/Nov/21
minimum value of function f(x)=(√((3sin x−4cos x−10)(3sin x+4cos x−10)))
$${minimum}\:{value}\:{of}\:{function}\: \\ $$$$\:\:\:{f}\left({x}\right)=\sqrt{\left(\mathrm{3sin}\:{x}−\mathrm{4cos}\:{x}−\mathrm{10}\right)\left(\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}−\mathrm{10}\right)} \\ $$
Commented by bobhans last updated on 19/Nov/21
g(x)=((3sin x−10)−4cos x)((3sin x−10)+4cos x) g(x)=(3sin x−10)^2 −16cos^2 x g(x)=9sin^2 x−60sin x+100−16(1−sin^2 x) g(x)=25sin^2 x−60sin x+84 g(x) = (5sin x−6)^2 +48 −5≤5sin x≤5 ⇒−11≤5sin x−6≤−1 ⇒0<(5sin x−6)^2 ≤121 g(x)_(min) = (5.1−6)^2 +48 = 49 f(x)_(min) =(√(49)) = 7
$$\:\mathrm{g}\left(\mathrm{x}\right)=\left(\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)−\mathrm{4cos}\:\mathrm{x}\right)\left(\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)+\mathrm{4cos}\:\mathrm{x}\right) \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)^{\mathrm{2}} −\mathrm{16cos}\:^{\mathrm{2}} \mathrm{x}\: \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{9sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{60sin}\:\mathrm{x}+\mathrm{100}−\mathrm{16}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right) \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{25sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{60sin}\:\mathrm{x}+\mathrm{84} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\:=\:\left(\mathrm{5sin}\:\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{48}\: \\ $$$$\:−\mathrm{5}\leqslant\mathrm{5sin}\:\mathrm{x}\leqslant\mathrm{5}\:\Rightarrow−\mathrm{11}\leqslant\mathrm{5sin}\:\mathrm{x}−\mathrm{6}\leqslant−\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}<\left(\mathrm{5sin}\:\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} \leqslant\mathrm{121} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)_{\mathrm{min}} \:=\:\left(\mathrm{5}.\mathrm{1}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{48}\:=\:\mathrm{49} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{min}} \:=\sqrt{\mathrm{49}}\:=\:\mathrm{7} \\ $$
Answered by mr W last updated on 19/Nov/21
f(x)=(√((3 sin x−10)^2 −(4 cos x)^2 )) f(x)=(√(9 sin^2 x+100−60 sin x−16 cos^2 x)) f(x)=(√(25 sin^2 x−60 sin x+84)) f(x)=(√(25(sin x−(6/( 5)))^2 +48)) f(x)_(min) =(√(25(1−(6/( 5)))^2 +48))=7 f(x)_(max) =(√(25(−1−(6/( 5)))^2 +48))=13
$${f}\left({x}\right)=\sqrt{\left(\mathrm{3}\:\mathrm{sin}\:{x}−\mathrm{10}\right)^{\mathrm{2}} −\left(\mathrm{4}\:\mathrm{cos}\:{x}\right)^{\mathrm{2}} } \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{9}\:\mathrm{sin}^{\mathrm{2}} \:{x}+\mathrm{100}−\mathrm{60}\:\mathrm{sin}\:{x}−\mathrm{16}\:\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{25}\:\mathrm{sin}^{\mathrm{2}} \:{x}−\mathrm{60}\:\mathrm{sin}\:{x}+\mathrm{84}} \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{25}\left(\mathrm{sin}\:{x}−\frac{\mathrm{6}}{\:\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{48}} \\ $$$${f}\left({x}\right)_{{min}} =\sqrt{\mathrm{25}\left(\mathrm{1}−\frac{\mathrm{6}}{\:\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{48}}=\mathrm{7} \\ $$$${f}\left({x}\right)_{{max}} =\sqrt{\mathrm{25}\left(−\mathrm{1}−\frac{\mathrm{6}}{\:\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{48}}=\mathrm{13} \\ $$
Commented by mnjuly1970 last updated on 19/Nov/21
bravo sir W
$${bravo}\:{sir}\:{W} \\ $$
Answered by FongXD last updated on 19/Nov/21
f(x)=(√((3sinx−10−4cosx)(3sinx−10+4cosx))) f(x)=(√((3sinx−10)^2 −(4cosx)^2 )) f(x)=(√(9sin^2 x−60sinx+100−16(1−sin^2 x))) f(x)=(√(25sin^2 x−60sinx+84)) f(x)=(√((5sinx−6)^2 +48)) Min of f(x) occurs when min of ∣5sinx−6∣ occurs we have: −1≤sinx≤1, ∀x∈R ⇔ −11≤5sinx−6≤−1 ⇔ 1≤∣5sinx−6∣≤11, ⇒ Min of ∣5sinx−6∣=1 therefore, Min of f(x)=(√(1^2 +48))=7
$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\left(\mathrm{3sinx}−\mathrm{10}−\mathrm{4cosx}\right)\left(\mathrm{3sinx}−\mathrm{10}+\mathrm{4cosx}\right)} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\left(\mathrm{3sinx}−\mathrm{10}\right)^{\mathrm{2}} −\left(\mathrm{4cosx}\right)^{\mathrm{2}} } \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{9sin}^{\mathrm{2}} \mathrm{x}−\mathrm{60sinx}+\mathrm{100}−\mathrm{16}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{25sin}^{\mathrm{2}} \mathrm{x}−\mathrm{60sinx}+\mathrm{84}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\left(\mathrm{5sinx}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{48}} \\ $$$$\mathrm{Min}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{occurs}\:\mathrm{when}\:\mathrm{min}\:\mathrm{of}\:\mid\mathrm{5sinx}−\mathrm{6}\mid\:\mathrm{occurs} \\ $$$$\mathrm{we}\:\mathrm{have}:\:−\mathrm{1}\leqslant\mathrm{sinx}\leqslant\mathrm{1},\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\Leftrightarrow\:−\mathrm{11}\leqslant\mathrm{5sinx}−\mathrm{6}\leqslant−\mathrm{1} \\ $$$$\Leftrightarrow\:\mathrm{1}\leqslant\mid\mathrm{5sinx}−\mathrm{6}\mid\leqslant\mathrm{11},\:\Rightarrow\:\mathrm{Min}\:\mathrm{of}\:\mid\mathrm{5sinx}−\mathrm{6}\mid=\mathrm{1} \\ $$$$\mathrm{therefore},\:\mathrm{Min}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{48}}=\mathrm{7} \\ $$
Commented by mnjuly1970 last updated on 19/Nov/21
Commented by mnjuly1970 last updated on 19/Nov/21
perfect
$${perfect} \\ $$

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