where are the vertical asymptotes on the graph of the secant function?\n$x = \\frac{\\pi}{2}+n\\pi$\n$x =…

where are the vertical asymptotes on the graph of the secant function?\n$x = \\frac{\\pi}{2}+n\\pi$\n$x = n\\pi$\n$x = 2n\\pi$\ncomplete\nthe range of the secant function is\n$y\\leq\\square$ or $y\\geq\\square$.

where are the vertical asymptotes on the graph of the secant function?\n$x = \\frac{\\pi}{2}+n\\pi$\n$x = n\\pi$\n$x = 2n\\pi$\ncomplete\nthe range of the secant function is\n$y\\leq\\square$ or $y\\geq\\square$.

Answer

Explanation:

Step1: Recall secant - cosine relation

The secant function is defined as $\sec(x)=\frac{1}{\cos(x)}$. Vertical asymptotes occur where the denominator $\cos(x) = 0$.

Step2: Find when cosine is zero

We know that $\cos(x)=0$ when $x=\frac{\pi}{2}+n\pi$, where $n\in\mathbb{Z}$ (the set of all integers).

Step3: Recall the range of secant

Since $\sec(x)=\frac{1}{\cos(x)}$, and the range of $\cos(x)$ is $[- 1,1]$. When $\cos(x)$ approaches $0$ from the positive - side, $\sec(x)\to+\infty$, and when $\cos(x)$ approaches $0$ from the negative - side, $\sec(x)\to-\infty$. Also, the maximum value of $\frac{1}{\cos(x)}$ occurs when $|\cos(x)|$ is minimum and non - zero, and the minimum value of $\frac{1}{\cos(x)}$ occurs when $|\cos(x)|$ is maximum. The range of $\cos(x)$ is $[-1,1]$. So the range of $\sec(x)$ is $y\leq - 1$ or $y\geq1$.

Answer:

For the vertical asymptotes: $x=\frac{\pi}{2}+n\pi$ For the range: $y\leq - 1$ or $y\geq1$