Answer
amplitude=1
period =$\displaystyle \frac{1}{2}$
graph (black):
.
Work Step by Step
See p. 424
For
$y=a\sin k(x-b),\ ,\quad y=a\cos k(x-b)$,
$(k>0)$
Amplitude: $|a|$, Period: $\displaystyle \frac{2\pi}{k}$,
Horizontal shift: $b$
------------------
$y=\cos 4\pi x$
$a=1 , k=4\pi, b=0$
So,
amplitude=$|1|=1$
period =$\displaystyle \frac{2\pi}{4\pi}=\frac{1}{2}$
To graph,
begin with $f(x)=\cos x,$ (red, dashed)
horizontally compress by factor $ 4\pi$ (black, solid line).
g(0)=$1$
$g(\displaystyle \frac{1}{8})=\cos(\frac{\pi}{2})=0$
$g(\displaystyle \frac{1}{4})=\cos(\pi)=-1$
$g(\displaystyle \frac{3}{8})=\cos(\frac{3\pi}{2})=0$
$g(\displaystyle \frac{1}{2})=\cos(2\pi)=1...$
