How do you find all local maximum and minimum points using the second derivative test given #y=2+3x-x^3#?

Question

1478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-04T09:57:09

maximum at (1,4) and minimum at (-1,0)

#y=2+3x-x^3#
#dy/dx = 3-3x^2#

the local may be maximum or minimum when
#dy/dx=0#

#0=3-3x^2#
#3x^2 = 3#
#x^2=1#
#x=1 and -1#.

when x = 1,
#y=2+3(1)-(1)^3 = 2+3-1 = 4#

#(d^2x) /dy^2=-6x#
# when x = 1, (dx^2)/dy^2 = -6(1) = -6#
since # (d^2x)/dy^2 < 0#, it is maximum point at (1,4)

when x = -1,
#y=2+3(-1)-(-1)^3 = 2-3+1 = 0#

#(d^2x) /dy^2=-6x#
# when x = -1, (dx^2)/dy^2 = -6(-1) = 6#
since # (d^2x)/dy^2 > 0#, it is minimum point at (-1,0)