graph exactly one cycle of the function $f(x)=18cos(2pi x)+9$. identify the maximum, minimum, and period of…

Explanation:

Step1: Recall cosine - function properties

The general form of a cosine function is $y = A\cos(Bx - C)+D$. For $f(x)=18\cos(2\pi x)+9$, we have $A = 18$, $B = 2\pi$, $C = 0$, $D=9$. The period of a cosine function is given by $T=\frac{2\pi}{|B|}$. Since $B = 2\pi$, $T=\frac{2\pi}{2\pi}=1$. The maximum value of $y = A\cos(Bx - C)+D$ is $A + D$ and the minimum is $-A+D$. Here, $A + D=18 + 9=27$ and $-A + D=-18 + 9=-9$.

Step2: Choose x - values for one - cycle

Since the period $T = 1$, we can choose $x = 0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1$ for one - cycle of the cosine function.

Step3: Calculate corresponding y - values

When $x = 0$: $f(0)=18\cos(2\pi\times0)+9=18\times1 + 9=27$ When $x=\frac{1}{4}$: $f(\frac{1}{4})=18\cos(2\pi\times\frac{1}{4})+9=18\cos(\frac{\pi}{2})+9=18\times0 + 9=9$ When $x=\frac{1}{2}$: $f(\frac{1}{2})=18\cos(2\pi\times\frac{1}{2})+9=18\cos(\pi)+9=18\times(-1)+9=-9$ When $x=\frac{3}{4}$: $f(\frac{3}{4})=18\cos(2\pi\times\frac{3}{4})+9=18\cos(\frac{3\pi}{2})+9=18\times0 + 9=9$ When $x = 1$: $f(1)=18\cos(2\pi\times1)+9=18\times1+9=27$

The five points are $(0,27),(\frac{1}{4},9),(\frac{1}{2},-9),(\frac{3}{4},9),(1,27)$.

Answer:

The five points for one - cycle of the function $f(x)=18\cos(2\pi x)+9$ are $(0,27),(\frac{1}{4},9),(\frac{1}{2},-9),(\frac{3}{4},9),(1,27)$