Practice problem
What is the range of the function f(x)=secxf(x)=\sec{x}f(x)=secx? Hint: secx=1cosx\sec{x}=\frac{1}{\cos{x}}secx=cosx1.
(−∞,−π2]∪[π2,∞)(-\infty, -\frac{\pi}{2}]\cup [\frac{\pi}{2}, \infty)(−∞,−2π]∪[2π,∞)
(−∞,∞)(-\infty, \infty)(−∞,∞)
[−1,1][-1, 1][−1,1]
(−∞,−1]∪[1,∞)(-\infty, -1]\cup [1, \infty)(−∞,−1]∪[1,∞)
Type a concept, like “slope” or “photosynthesis”.