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)#