To find the local minimum, local maximum, and the concavity is to get the first and second derivative of the function:
Given: y=sin x on [0,2π]
(1) For local min and max, get first derivative:
y'=cos x
Then set y'=0,
cos x = 0
The solution for x on [0,2π]:
x= π/2, 3π/2
Use the original equation to get the coordinates of local min and max:
y=sin (π/2), y=sin (3π/2)
y=1, y=-1
∴Local min is at point (π/2,1) and
Local max is at point (3π/2,-1)
y'' = -sin x
To find the concavity, you need to determine the point of inflection and that is:
y'' = 0
0=-sin x
x= 0, π, 2π
∴ The point of inflection (concavity) is at the point (π,0). We can't use x=0, 2π because they are not differentiable on the given interval [0,2π].
From point (0,0)→ (π/2,1)→ (π,0), the concavity is open down.
From point (π,0)→ (3π/2,-1)→ (2π,0), the concavity is open up.
