find de maximun and minumun points of f(x)=x^2/x+2
f = x^2/(x + 2)
df/dx = ((x + 2).2x – x^2.1) / (x + 2)
df/dx = (2x^2 + 4x – x^2) / (x + 2)
df/dx = x(x + 4)/(x + 2)
when df/dx = 0, x = 0, x = -4
There are turning points at x =0, and at x = -4.
x = 0
f(0) = 0
Maximum is at (0, 0) and has the value Max = 0
x = -4
f(-4) = (-4)^2 / (-4 + 2) = 16/(-2) = -8
f(-4) = -8
Minimum is at (-4, -8) and has the value Min =-8