Question

Encontre o valor máximo da função, f(x) = 3cos x + 2sen x





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Por favor responder de forma detalhada. Respostas com brincadeiras serão eliminadas.

Answer 1

$\mathrm{f(x)=3\cos{x}+2\sin{x}}\\\\ \textrm{Os pontos cr\'iticos de uma fun\c{c}\~ao s\~ao dados quando}\\ \textrm{igualamos sua derivada primeira a zero.}\\\\ \mathrm{f'(x)=3\dfrac{d}{dx}(\cos{x})+2\dfrac{d}{dx}(\sin{x})}\\\\ \mathrm{f'(x)=3(-\sin{x})+2\cos{x}\ \to\ f'(x)=2\cos{x}-3\sin{x}}\\\\ \mathrm{0=2\cos{x}-3\sin{x}\ \to\ 3\sin{x}=2\cos{x}}\\\\ \mathrm{\dfrac{\sin{x}}{\cos{x}}=\dfrac{2}{3}\ \to\ \tan{x}=\dfrac{2}{3}\ \to\ x=\arctan{\bigg(\dfrac{2}{3}\bigg)+\pi k}}$

$\mathrm{M\'aximos\ \to\ x=\arctan{\bigg(\dfrac{2}{3}\bigg)}+2\pi k}\\\\ \mathrm{M\'inimos\ \to\ x=\arctan{\bigg(\dfrac{2}{3}\bigg)}+2\pi k+\pi}}\\\\\\ \mathbf{Valor\ m\'aximo\ de\ f(x):}\\\\ \mathrm{f(x)=3\cos{\bigg(\arctan{\bigg(\dfrac{2}{3}\bigg)}\bigg)}+2\sin{\bigg(\arctan{\bigg(\dfrac{2}{3}\bigg)}\bigg)}}\\\\ \mathrm{Lembrando\ que:}\\\\ \mathrm{\sin{(\arctan{x})}=\dfrac{x\sqrt{1+x^2}}{1+x^2}\ \ \| \ \ \cos{(\arctan{x})}=\dfrac{\sqrt{1+x^2}}{1+x^2}}$

$\mathrm{f(x)=3\Bigg(\dfrac{\sqrt{1+\big(\frac{2}{3}\big)^2}}{1+\big(\frac{2}{3}\big)^2}\Bigg)+2\Bigg(\dfrac{\frac{2}{3}\sqrt{1+\big(\frac{2}{3}\big)^2}}{1+\big(\frac{2}{3}\big)^2}\Bigg)}\\\\\\ \mathrm{f(x)=3\Bigg(\dfrac{\sqrt{\frac{9}{9}+\frac{4}{9}}}{\frac{9}{9}+\frac{4}{9}}\Bigg)+\dfrac{4}{3}\Bigg(\dfrac{\sqrt{\frac{9}{9}+\frac{4}{9}}}{\frac{9}{9}+\frac{4}{9}}\Bigg)}$

$\mathrm{f(x)=\bigg(\dfrac{9}{3}+\dfrac{4}{3}\bigg)\Bigg(\dfrac{\sqrt{\frac{13}{9}}}{\frac{13}{9}}\Bigg)\ \to\ f(x)=\dfrac{13}{3}.\dfrac{\sqrt{13}}{3}.\dfrac{9}{13}}\\\\\\ \mathrm{f(x)=\dfrac{9\sqrt{13}}{9}\ \to\ \mathbf{f(x)=\sqrt{13}}}$