Let f(x) = x^2(x-5)(x-4)(x-3)(x+3)(x+4)(x+5). What are the value(s) of x at which f(x) attains its minimum? What is this minimum value?

Level: Senior

Setting y = x^4-25x^2+72,
f(x)
= x^2*[(x-3)(x+3)][(x-4)(x+4)][(x-5)(x+5)]
= x^2(x^2-9)(x^2-16)(x^2-25)
= [x^2(x^2-25)][(x2-9)(x^2-16)]
= (x^4-25x^2)(x^4-25x^2+144)
= (y-72)(y+72)
= y^2-5184.
Since y^2 is minimum if y = 0, so f(x) is minimum if y = 0 has a solution in x. Solving y = 0 for x:
x^4-25x^2+72 = 0,
x^2 = [25+-sqrt((-25)^2-4*72)]/2 = [25+-sqrt(337)]/2 ~ +-21.68 or +-3.32,
x ~ +-4.66 or +-1.82
Since y = 0 at x ~ +-4.66 or +-1.82, so the values of x at which f(x) attains its minimum 0^2-5184 = -5184 are +-4.66 and +-1.82.