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Question Number 128753 by bramlexs22 last updated on 10/Jan/21

Find maximum and minimum value of function f(x)=27^(cos 2x) . 81^(sin 2x) ?

$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{27}^{\mathrm{cos}\:\mathrm{2x}} .\:\mathrm{81}^{\mathrm{sin}\:\mathrm{2x}} \:? \\ $$

Answered by liberty last updated on 10/Jan/21

⇔ f(x) = 3^(3cos 2x) . 3^(4sin 2x) f(x)=3^(3cos 2x+4sin 2x) = 3^(5cos (2x−α)) we know that −1≤ cos (2x−α)≤1 it follows that max f(x)= 3^(5.1) = 3^5 =243 and minimum f(x)=3^(5.(−1)) =3^(−5) = (1/(243)) .

$$\:\Leftrightarrow\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{3}^{\mathrm{3cos}\:\mathrm{2x}} .\:\mathrm{3}^{\mathrm{4sin}\:\mathrm{2x}} \\ $$$$\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3}^{\mathrm{3cos}\:\mathrm{2x}+\mathrm{4sin}\:\mathrm{2x}} \:=\:\mathrm{3}^{\mathrm{5cos}\:\left(\mathrm{2x}−\alpha\right)} \\ $$$$\:\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:−\mathrm{1}\leqslant\:\mathrm{cos}\:\left(\mathrm{2x}−\alpha\right)\leqslant\mathrm{1}\:\mathrm{it}\:\mathrm{follows}\: \\ $$$$\mathrm{that}\:\mathrm{max}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{3}^{\mathrm{5}.\mathrm{1}} \:=\:\mathrm{3}^{\mathrm{5}} =\mathrm{243}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3}^{\mathrm{5}.\left(−\mathrm{1}\right)} =\mathrm{3}^{−\mathrm{5}} \:=\:\frac{\mathrm{1}}{\mathrm{243}}\:. \\ $$

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