Given: #y = 2x^4 -4x^2 + 1#
Infection points can be found by setting #y'' = 0#; but first you need to know if you have minimums and/or maximums. You can use the first derivative test or the second derivative test. The second derivative test is easier if the second derivative is easy to find. However, it has a restriction. If #y''(critical" "value)" = 0# you will then need to use the first derivative test.
Let's use the second derivative test:
Find the first derivative: #y' = 8x^3 - 8x = 0#
Factor: #8x(x^2-1) = 8x(x - 1)(x + 1) = 0#
critical values: #x = 0, x = +-1#
Find the second derivative: #y'' = 24x^2 -8#
Find the relative minimums and maximums:
# y''(0) < 0# relative maximum at #x = 0#
#y''(-1) > 0# relative minimum at #x = -1#
#y''(1) > 0# relative minimum at #x = 1#
Find the points of inflection:
#y'' = 24x^2 -8 = 0#
Factor: #8(3x^2 - 1) = 0#
#3x^2 = 1#
#x^2 = 1/3#
#x = +- sqrt(1/3) = +- 1/(sqrt(3)) = +- (sqrt(3))/3#
#y((sqrt(3))/3) = 2((sqrt(3))/3)^4 - 4((sqrt(3))/3)^2 +1 = -1/9#
#y(-(sqrt(3))/3) = 2((sqrt(3))/3)^4 - 4((sqrt(3))/3)^2 +1 = -1/9#
inflection points #(sqrt(3)/3, -1/9), (-sqrt(3)/3, -1/9)#
Concave down is when #y'# is decreasing. A relative maximum would be concave down to the point of inflection.
Concave up is when #y'# is increasing. A relative minimum would be concave up to the point of inflection.
Intervals of concavity:
#(-oo, -sqrt(3)/3), (-sqrt(3)/3. sqrt(3)/3), (sqrt(3)/3, oo)#
remember from work above, that there is a relative maximum at #x = 0# and two relative minimums at #x = +-1#
concave up #(-oo, -sqrt(3)/3), (sqrt(3)/3, oo)#
concave down #(-sqrt(3)/3, sqrt(3)/3)#
