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For what values of x is #f(x)= -x^3+3x^2+2x-12 # concave or convex?

1 Answer
Jan 4, 2017

Answer:

As viewed from O, concave #( y' uarr)# for #x < 1# and concave #(y' darr)# for# x >1. (1, -8)# is the point of inflexion (POI), at which the tangent crosses the curve , for reversing rotation.

Explanation:

#y=f(x)=-x^3+3x^2+2x-12#

#y'=-3x^2+6x+2 =0#, at the turning points#x =1+-sqrt(5/3)=-0.291 and 2.291#, nearly

y''--6(x-1)=0#, at x =1.

y'''=-6 #ne 0#

So, x =1 gives the point of inflexion (POI) #(1. -8)#.

Here, the tangent crosses the curve, reversing rotation, from

anticlockwise to clockwise.

The second graph, the zooming is to see #POI (1. -8)# in #Q_4#,

at the level #y = - 8#.

graph{-x^3+3x^2+2x-12 [-29.95, 29.95, -14.97, 14.98]}

graph{(-x^3+3x^2+2x-12-y)(y+8)=0 [-2, 2, -20, 20]} .