Respuesta :
The given function is
[tex]f(x)=4\cos \text{(}\pi x)+1[/tex]The general form of the cosine function is
[tex]y=a\cos (bx+c)+d[/tex]a is the amplitude
2pi/b is the period
c is the phase shift
d is the vertical shift
By comparing the two functions
a = 4
b = pi
c = 0
d = 1
Then its period is
[tex]\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}[/tex]The equation of the midline is
[tex]y_{ml}=\frac{y_{\max }+y_{\min }}{2}[/tex]Since the maximum is at the greatest value of cos, which is 1, then
[tex]\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}[/tex]Since the minimum is at the smallest value of cos, which is -1, then
[tex]\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}[/tex]Then substitute them in the equation of the midline
[tex]\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}[/tex]The answers are:
Period = 2
Equation of the midline is y = 1
Maximum = 5
Minimum = -3
