which of the following is true for f(x) = 5cos(x) +1?\nthe period of the function is 10π.\nthe function has…

Explanation:

Step1: Recall the general form of a cosine - function

The general form of a cosine function is $y = A\cos(Bx - C)+D$, where $A$ is the amplitude, $B$ affects the period ($T=\frac{2\pi}{|B|}$), $C$ is the phase - shift, and $D$ is the vertical shift. For the function $f(x)=5\cos(x)+1$, we have $A = 5$, $B = 1$, $C = 0$, and $D = 1$.

Step2: Calculate the period

Using the formula $T=\frac{2\pi}{|B|}$, with $B = 1$, we get $T=\frac{2\pi}{|1|}=2\pi\neq10\pi$.

Step3: Calculate the amplitude

The amplitude $|A|$. Since $A = 5$, the amplitude is $|5| = 5\neq2.5$.

Step4: Check for a zero

Set $f(x)=0$, so $5\cos(x)+1 = 0$. Then $\cos(x)=-\frac{1}{5}$. When $x=\frac{\pi}{2}$, $\cos(\frac{\pi}{2}) = 0$, and $f(\frac{\pi}{2})=5\times0 + 1=1\neq0$. So $(\frac{\pi}{2},0)$ is not a zero of the function.

Step5: Find the range

The range of the cosine function $y = \cos(x)$ is $[- 1,1]$. For the function $y = 5\cos(x)+1$, when $\cos(x)=-1$, $y=5\times(-1)+1=-4$; when $\cos(x)=1$, $y=5\times1 + 1=6$. So the range of $y = 5\cos(x)+1$ is $-4\leq y\leq6$.

Answer:

The range of the function is the set of real numbers $-4\leq y\leq6$.