Final answer:
The graph of the function y=5+2sec(x)/sec(x) has a horizontal tangent line at x = π/2 in the interval 0≤x≤2π.
Step-by-step explanation:
The function y=5+2sec(x)/sec(x) has horizontal tangent lines when its derivative is equal to zero. Let's find the derivative of the given function:
The derivative of y=5+2sec(x)/sec(x) can be found using the quotient rule and chain rule. The derivative is given by dy/dx = (2sec(x)tan(x)sec(x) - 2sec^2(x))/sec^2(x). Simplifying this expression, we have dy/dx = (2tan(x) - 2sec(x))/sec^3(x).
To find the horizontal tangent lines, we need to solve the equation dy/dx = 0. Setting the derivative equal to zero, we have (2tan(x) - 2sec(x))/sec^3(x) = 0. Simplifying this equation, we get 2tan(x) - 2sec(x) = 0. Rearranging the terms, we have tan(x) = sec(x).
Since tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x), we can rewrite the equation as sin(x)/cos(x) = 1/cos(x). Multiplying both sides by cos(x), we have sin(x) = 1. This equation is satisfied when x = π/2. Therefore, the graph of the function has a horizontal tangent line at x = π/2 in the interval 0≤x≤2π.