Intervals of increase/decrease of secx.

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SUMMARY

The discussion focuses on determining the intervals of increase and decrease for the function sec(x) on the interval (-π/2, 3π/2). The derivative f'(x) = sin(x)/cos²(x) was calculated, leading to critical points at x = 0 and x = π. The analysis of the derivative at test points within the intervals (-π/2, 0), (0, π), and (π, 3π/2) reveals that sec(x) decreases from -π/2 to 0 and from π to 3π/2, while it increases from 0 to π, with a vertical asymptote at x = π/2 where sec(x) is undefined.

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  • Understanding of calculus, specifically derivatives
  • Knowledge of trigonometric functions, particularly secant
  • Familiarity with critical points and their significance
  • Basic concepts of the unit circle and its application in trigonometry
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  • Study the properties of the secant function and its behavior near asymptotes
  • Learn how to analyze functions using the first derivative test
  • Explore the concept of vertical asymptotes in trigonometric functions
  • Practice finding intervals of increase and decrease for other trigonometric functions
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Homework Statement



Find the intervals of increase and decrease of secx on the interval (-pi/2, 3pi/2).

Homework Equations





The Attempt at a Solution



I found the derivative and set it equal to zero to get the critical points:
f'(x)=sinx/cos^2x
0=sinx/cos^2x
0=sinx
x= 0 and pi (restricted by interval)

Now to find the intervals of increase and decrease I am lost? It says think about the unit circle?
 
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dylanhouse said:
I found the derivative and set it equal to zero to get the critical points:
f'(x)=sinx/cos^2x
0=sinx/cos^2x
0=sinx
x= 0 and pi (restricted by interval)

Now to find the intervals of increase and decrease I am lost? It says think about the unit circle?

Now pick a point in each of the following intervals: (-\pi/2, 0), (0, \pi), (\pi, 3\pi/2) and evaluate the derivative at those points. How can you use this to determine where the function is increasing or decreasing?

Also, note that there is one point in (-\pi/2, 3\pi/2) where \sec(x) is undefined. So, in particular, the function can't be increasing or decreasing at that point.
 
I evaluated -pi/4 < 0, pi/2 >0 and 5pi/4 < 0. This would mean a decrease from -pi/2 to 0 and pi to 3pi/2. And an increase from 0 to pi?
 
Except at pi/2 where there is a vertical asymptote.
 

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