Let $f ( x ) = \sec x$ for $x \in [ 0,2 \pi ]$ The set of all vertical asymptotes of ƒ is

A) $\left\{ x = \frac { \pi } { 2 } \right\}$
B) $\left\{ x = \frac { 3 \pi } { 2 } \right\}$
C) $\left\{ x = - \frac { \pi } { 2 } \right\}$
D) $\left\{ x = \frac { \pi } { 2 } , x = \frac { 3 \pi } { 2 } \right\}$
E) $\left\{ x = - \frac { \pi } { 2 } , x = \frac { 3 \pi } { 2 } \right\}$

Question

Let $f ( x ) = \sec x$ for $x \in [ 0,2 \pi ]$ The set of all vertical asymptotes of ƒ is

A) $\left\{ x = \frac { \pi } { 2 } \right\}$
B) $\left\{ x = \frac { 3 \pi } { 2 } \right\}$
C) $\left\{ x = - \frac { \pi } { 2 } \right\}$
D) $\left\{ x = \frac { \pi } { 2 } , x = \frac { 3 \pi } { 2 } \right\}$
E) $\left\{ x = - \frac { \pi } { 2 } , x = \frac { 3 \pi } { 2 } \right\}$

Quizplus · Accepted Answer

The secant function, $\sec x$, has vertical asymptotes where the cosine function, $\cos x$, is equal to zero since $\sec x = \frac{1}{\cos x}$. Within the interval $[0, 2\pi]$, $\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. Therefore, the vertical asymptotes of $f(x) = \sec x$ are at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

Let f(x)=sec⁡xf ( x ) = \sec xf(x)=secx For x∈[0,2π]x \in [ 0,2 \pi ]x∈[0,2π]

Let $f ( x ) = \sec x$ For $x \in [ 0,2 \pi ]$