question\ngraph exactly one cycle of the function $f(x)=14cos(\frac{1}{5}x)$.\nidentify the maximum…

Explanation:

Step1: Recall cosine - function properties

The general form of a cosine function is $y = A\cos(Bx - C)+D$. For the function $f(x)=14\cos(\frac{1}{5}x)$, we have $A = 14$, $B=\frac{1}{5}$, $C = 0$, and $D = 0$.

Step2: Find the maximum value

The range of the basic cosine function $y=\cos(t)$ is $[- 1,1]$. When we have $y = A\cos(Bx - C)+D$, the maximum value occurs when $\cos(Bx - C)=1$. For $y = 14\cos(\frac{1}{5}x)$, when $\cos(\frac{1}{5}x)=1$, $y_{max}=14\times1=14$.

Step3: Find the minimum value

The minimum value of the basic cosine function $y = \cos(t)$ is $-1$. For $y = 14\cos(\frac{1}{5}x)$, when $\cos(\frac{1}{5}x)=-1$, $y_{min}=14\times(-1)=-14$.

Step4: Find the period

The period of a cosine function $y = A\cos(Bx - C)+D$ is given by the formula $T=\frac{2\pi}{|B|}$. Since $B=\frac{1}{5}$, then $T=\frac{2\pi}{\frac{1}{5}}=10\pi$.

Answer:

Maximum: 14 Minimum: - 14 Period: $10\pi$