Find the local maximum and minimum values of the following curve within the given domain.
y=SECx-TANx 0 ≤ x ≤ 2π
y = secx - tanx
y' = secx(tanx-secx)
y' = 0 when
secx=0 -- never
tanx-secx=0
secx(sinx-1) = 0
secx=0 -- never
sinx=1 -- x = pi/2
Now, we want x=pi/2 to be a solution, but unfortunately, neither secx nor tanx is defined there.
There are no min/max points on the graph of f(x), as seen at this url:
http://www.wolframalpha.com/input/?i=secx+-+tanx
To find the local maximum and minimum values of the curve y = sec(x) - tan(x) within the domain 0 ≤ x ≤ 2π, we need to follow these steps:
Step 1: Find the critical points.
To find the critical points, we need to find where the derivative of the function is equal to zero or undefined. In this case, the derivative of y with respect to x is:
dy/dx = sec(x)tan(x) + sec^2(x)
Setting the derivative equal to zero, we have:
sec(x)tan(x) + sec^2(x) = 0
Divide both sides by sec(x) to simplify:
tan(x) + sec(x) = 0
Using the identity sec(x) = 1/cos(x) and rearranging the equation, we get:
sin(x) + 1 = 0
sin(x) = -1
Solving for x within the domain 0 ≤ x ≤ 2π, we have:
x = π/2, 3π/2
Step 2: Evaluate the function at the critical points and the endpoints of the domain.
Evaluate y = sec(x) - tan(x) at the critical points and the endpoints of the domain:
At x = 0, y = sec(0) - tan(0) = 1 - 0 = 1
At x = π/2, y = sec(π/2) - tan(π/2) = undefined (division by zero)
At x = 2π, y = sec(2π) - tan(2π) = 1 - 0 = 1
At x = 3π/2, y = sec(3π/2) - tan(3π/2) = undefined (division by zero)
Step 3: Determine the local maximum and minimum values.
Based on the above evaluations, we have the following values:
At x = 0, y = 1 (local maximum)
At x = 2π, y = 1 (local minimum)
Therefore, the local maximum value is 1 and the local minimum value is also 1 for the curve y = sec(x) - tan(x) within the domain 0 ≤ x ≤ 2π.