Question

What is the range of the function $g(x) = sec(x)$?

• A. (0, $\infty$)
• B. (-$\infty$,$\infty$)
• C. (-1,1)
• D. (-$\infty$,-1]$\cup$[1,$\infty$)

Answer 1

Answer: (D)

(-$\infty$, -1] $\cup$ [1, $\infty$)

Answer 2

The range of a function refers to the set of all possible output values (y-values) that the function can produce.

The given function is $g(x) = \sec(x)$. We know that $\sec(x)$ is defined as $\frac{1}{\cos(x)}$.

The range of the cosine function, $\cos(x)$, is $[-1, 1]$. This means that for any real number $x$, $-1 \le \cos(x) \le 1$.

Now, let's consider the values of $\sec(x) = \frac{1}{\cos(x)}$ based on the values of $\cos(x)$:

1. When $\cos(x)$ is positive: If $0 < \cos(x) \le 1$: As $\cos(x)$ approaches 0 from the positive side, $\frac{1}{\cos(x)}$ approaches $+\infty$. When $\cos(x) = 1$, $\frac{1}{\cos(x)} = \frac{1}{1} = 1$. So, for $0 < \cos(x) \le 1$, the values of $\sec(x)$ are in the interval $[1, \infty)$.

2. When $\cos(x)$ is negative: If $-1 \le \cos(x) < 0$: As $\cos(x)$ approaches 0 from the negative side, $\frac{1}{\cos(x)}$ approaches $-\infty$. When $\cos(x) = -1$, $\frac{1}{\cos(x)} = \frac{1}{-1} = -1$. So, for $-1 \le \cos(x) < 0$, the values of $\sec(x)$ are in the interval $(-\infty, -1]$.

Combining these two cases, the range of $\sec(x)$ includes all real numbers except those strictly between -1 and 1. Therefore, the range of $g(x) = \sec(x)$ is $(-\infty, -1] \cup [1, \infty)$.