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Fie f : ( -pi/2 , pi/2 ) -> R, f(x)=sinx+√3cosx. Sa se determine imaginea functiei f

Răspuns :

[tex]E = \sin x+\sqrt3 \cos x \\ \\\dfrac{E}{2} = \dfrac{1}{2}\sin x+\dfrac{\sqrt 3}{2}\cos x\\ \\ \dfrac{E}{2} = \cos\dfrac{\pi}{3}\sin x + \sin \dfrac{\pi}{3}\cos x \\ \\ \dfrac{E}{2} = \sin\Big(x+\dfrac{\pi}{3}\Big)\\ \\ E = 2\sin \Big(x+\dfrac{\pi}{3}\Big) \\ \\\\ f(x)=2\sin \Big(x+\dfrac{\pi}{3}\Big) \\ \\ f'(x) = 2\cos \Big(x+\dfrac{\pi}{3}\Big)= 0 \Rightarrow x+\dfrac{\pi}{3}= \dfrac{(2k+1)\pi}{2} \Rightarrow[/tex]

[tex]\Rightarrow x =\dfrac{\pi}{6}\\ \\\\\text{Extremele sunt:}\\\\ f\Big(-\dfrac{\pi}{2}\Big) = -1,\quad f\Big(\dfrac{\pi}{6}\Big) = 2,\quad f\Big(\dfrac{\pi}{2}\Big) = 1 \\ \\ \\\Rightarrow \boxed{Imf = (-1,2]}[/tex]