Answer:
The required function is [tex]f(x)=2\cos (\frac{\pi}{3}x-\frac{2\pi}{3})[/tex].
Step-by-step explanation:
Note: The value of period and phase shift are not given properly.
Consider amplitude = 2, period = 6, phase shift = 2.
The general form of cosine function is
[tex]f(x)=A\cos (Bx-C)+D[/tex] ..... (1)
where, |A| is amplitude, [tex]\frac{2\pi}{B}[/tex] is period, C/B is phase shift and D is midline.
From the given information we conclude that
[tex]|A|=2[/tex]
[tex]\frac{2\pi}{B}=6[/tex]
[tex]\frac{2\pi}{6}=B[/tex]
[tex]\frac{\pi}{3}=B[/tex]
[tex]\text{Phase shift}=\frac{C}{B}[/tex]
[tex]\text{Phase shift}\times B=C[/tex]
[tex]2\times \frac{\pi}{3}=C[/tex]
[tex]\frac{2\pi}{3}=C[/tex]
Substitute A=2, [tex]B=\frac{\pi}{3}[/tex] and [tex]C=\frac{2\pi}{3}[/tex] and D=0 in equation (1).
[tex]f(x)=2\cos (\frac{\pi}{3}x-\frac{2\pi}{3})+0[/tex]
[tex]f(x)=2\cos (\frac{\pi}{3}x-\frac{2\pi}{3})[/tex]
Therefore, the required function is [tex]f(x)=2\cos (\frac{\pi}{3}x-\frac{2\pi}{3})[/tex].
