## Precalculus (6th Edition) Blitzer

Consider the provided function, $y=\frac{1}{2}\sec 2\pi x$ By use the reciprocal function, $y=\frac{1}{2}\cos 2\pi x$ Now substitute different values of $x$ in the provided function to find the values of $y$. As the interval is given, put values of $x$ in the range of the interval, $\text{0}\le x\le 2$ To find the key points of the function, put $x=0$ in the provided function, \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot 0 \right) \\ & =\frac{1}{2}\cdot 1 \\ & =\frac{1}{2} \end{align} And, put $x=\frac{1}{4}$ \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{1}{4} \right) \\ & =\frac{1}{2}\cdot 0 \\ & =0 \end{align} And, put $x=\frac{1}{2}$ \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{1}{2} \right) \\ & =\frac{1}{2}\cdot \left( -1 \right) \\ & =-\frac{1}{2} \end{align} And, put $x=\frac{3}{4}$ \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{3}{4} \right) \\ & =\frac{1}{2}\cdot 0 \\ & =0 \end{align} And, put $x=1$ \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot 1 \right) \\ & =\frac{1}{2}\cdot 1 \\ & =\frac{1}{2} \end{align} And, put $x=\frac{5}{4}$ \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{5}{4} \right) \\ & =\frac{1}{2}\cdot \left( 0 \right) \\ & =0 \end{align} And, put $x=\frac{3}{2}$ \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{3}{2} \right) \\ & =\frac{1}{2}\cdot \left( -1 \right) \\ & =-\frac{1}{2} \end{align} And, put $x=\frac{7}{4}$ \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{7}{4} \right) \\ & =\frac{1}{2}\cdot \left( 0 \right) \\ & =0 \end{align} And, put $x=2$ \begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot 2 \right) \\ & =\frac{1}{2}\cdot \left( 1 \right) \\ & =\frac{1}{2} \end{align}