Use the sine/cosine quotient or reciprocal definition to construct \(y=\sec x\) on \((- \pi,\pi)\). State its real domain, least period, parity, full pole set, range, zeros or vertices, and both one-sided behaviors at the first nonnegative pole. Evaluate it at \(\dfrac{\pi}{5}\) where defined. Explain why no finite sinusoidal amplitude applies. Curve 1: \(\sec{\left(x \right)}\). Show the requested calculations and reasoning. Written explanations and sketches require teacher review; choosing a summary does not verify that work. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.
Answer: Least period \(2 \pi\); parity even; poles \(x=\dfrac{\pi}{2}+n\pi\), \(n\in\mathbb{Z}\).
Use \(\sec x=\dfrac1{\cos x}\). Denominator \(\ne0\), so exclude \(x=\dfrac{\pi}{2}+n\pi\). The parity is even from the sine/cosine parity rules. The least period is \(2 \pi\). The zero set of tangent/cotangent or the alternating sign of consecutive reciprocal branches rules out half this period. The range is \((-\infty,-1]\cup[1,\infty)\). There are no zeros. Vertices are at \((n\pi,(-1)^n)\) for secant, or \((\dfrac\pi2+n\pi,(-1)^n)\) for cosecant; use the formula for the selected function. At the first nonnegative pole \(\dfrac{\pi}{2}\), the left limit is positive infinity and the right limit is negative infinity. Check the denominator sign on each side; do not connect the branches. The value at \(\dfrac{\pi}{5}\) is \(-1 + \sqrt{5}\). These functions are unbounded. A vertical scale factor is not a finite amplitude. Written-work review:
Use the defining quotient or reciprocal to derive the entire domain and pole set.
Give least period, parity, range and every zero or vertex family.
Draw disconnected branches, justify both one-sided pole signs and evaluate the specified input.