What is the turning point on the curve f(x) = 2x^2 - 2x + 4
df/dx = 4x - 2
turning point when differential = 0
==>
4x = 2 hence; x = 0.5
When x = 0.5 f(x) is equal to (substitute) 3.5
hence the turning point is at (0.5,3.5)
DT
df/dx = 4x - 2
turning point when differential = 0
==>
4x = 2 hence; x = 0.5
When x = 0.5 f(x) is equal to (substitute) 3.5
hence the turning point is at (0.5,3.5)