Question

For what value(s) of b, if any, will the function q(x) = (x^4)/2 +b*x^2 +32x-5 have a derivative of zero AND a point of inflection at the same value of x?

Answer 1

q ' (x) = 2x^3 + 2bx + 32
q '' (x) = 6x^2 + 2b

both are equal to zero, so

6x^2 + 2b = 0
2b = -6x^2
b = -3x^2

and

2x^3 + 2bx + 32 = 0
2x^3 + 2(-3x^2)(x) + 32 = 0
-4x^3 = -32
x^3 = 8
x = 2

then b = -3(4) = -12

check:
q(x) = (1/2)x^4 - 12x^2 + 32x - 5
q ' (x) = 2x^3 - 24x + 32
= 0
x^3 - 12x + 16 = 0
(x-2)(x^2 + 2x - 8) = 0
(x-2)(x+4)(x-2) = 0
so g ' (x) is zero when x = 2
q '' (x) = 6x^2 - 24
at points of inflection,
6x^2 - 24 = 0
x^2 - 4 = 0
(x-2)(x+2) = 0
x = 2 produces a point of inflection
so x=2 works for both conditions and needs
b = 12 to be true.