Graph natural log of sec(x)

Problem

y=ln(sec(x))

Solution

1. Identify the domain of the function. The natural logarithm requires its argument to be positive, so sec(x)>0 This occurs when cos(x)>0 which corresponds to the intervals (−π/2+2*π*k,π/2+2*π*k) for any integer k

2. Determine the period and symmetry. Since sec(x) has a period of 2*π the function y=ln(sec(x)) also has a period of 2*π Additionally, because sec(−x)=sec(x) the function is even, meaning it is symmetric about the yaxis.

3. Find vertical asymptotes. As x approaches π/2 or −π/2 from within the domain, sec(x) approaches ∞ so ln(sec(x)) approaches ∞ There are vertical asymptotes at x=π/2+π*k

4. Locate the intercepts. The yintercept occurs at x=0 where y=ln(sec(0))=ln(1)=0 Since sec(x)≥1 for all x in the domain, ln(sec(x))≥0 meaning the xintercepts are the local minima at x=2*π*k

5. Analyze the derivative to find the slope. Using the chain rule:

d(y)/d(x)=1/sec(x)⋅sec(x)*tan(x)=tan(x)

The function increases when tan(x)>0 and decreases when tan(x)<0

6. Sketch the graph. Draw U-shaped curves opening upward between the vertical asymptotes, with the lowest points (vertices) touching the xaxis at multiples of 2*π

Final Answer

y=ln(sec(x))

Want more problems? Check here!