Question

In Exercises \(27-34,\) (a) find the vertical asymptotes of the graph of \(f(x) .(\) b) Describe the behavior of \(f(x)\) to the left and right of each vertical asymptote. $$f(x)=\sec x$$

Short answer

The vertical asymptotes of the secant function are at \(x = (2n+1)\pi/2\), where n is any integer. On the intervals \((-2n\pi -\pi/2, -2n\pi +\pi/2)\) and \((2n\pi +\pi/2, 2n\pi +3\pi/2)\), the secant function approaches positive infinity, and on the intervals \((-2n\pi +\pi/2, -2n\pi +3\pi/2)\) and \((2n\pi -\pi/2, 2n\pi +\pi/2)\), the secant function approaches negative infinity.

Step-by-step solution

Identify the secant function

The secant function, denoted as \(\sec x\), is the reciprocal of the cosine function. It is defined as \(\sec x = 1/ \cos x\).

Determine the vertical asymptotes

The vertical asymptotes of the secant function are located where the cosine function equals to zero. This occurs at \(x = (2n+1)\frac{\pi}{2}\), where n is any integer.

Describe the behavior

On the intervals \((-2n\pi -\pi/2, -2n\pi +\pi/2)\) and \((2n\pi +\pi/2, 2n\pi +3\pi/2)\), the secant function increases without bound (i.e., approaches positive infinity). On the intervals \((-2n\pi +\pi/2, -2n\pi +3\pi/2)\) and \((2n\pi -\pi/2, 2n\pi +\pi/2)\), the secant function decreases without bound (i.e., approaches negative infinity). This is the behavior to the left and right of each vertical asymptote.