f'(x) = 6x^2 - 6x - 12
= 0 for a max/min (critical value)
divide by 6
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
so, yes, x=2 is a critical number, (the other is x = -1)
f '' (x) = 12x-6
f(2) = 2(2^3) - 3(2^2) - 12(2) + 18 = -2
f '' (2) = 24-6 > 0
since at (2,-2) , the value of f '' (2) is positive, the curve is opening upwards at that point and thus
(2,-2) is a minimum point.
http://www.wolframalpha.com/input/?i=plot+y+%3D++2x%5E3-3x%5E2-12x%2B18
I just love this Wolfram website
Decide if the given value of x is a critical number for f, and if so, decide whether the point for x on f is a relative minimum, relative maximum, or neither.
f(x)= 2x^3-3x^2-12x+18; x=2
the ans. is "Critical number, minimum at (2,-2) but don't know how?
1 answer