An absolute minimum for f(x) = 2x^3 - 3x^2 -12x - 4 occurs______ Possible solutions below?

Question

These are the possible solutions

where x = 1 .

where x = 2 .

where x = 3 .

for a value of x which is not listed above.

for no value of x

Answer 1

f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2)
f''(x) = 6(2x-1).

f'(x) = 0 when x^2-x-2=0.
Factor it and you get (x+1)(x-2)=0, so x=-1 and x=2 are the critical points.

f''(2) = 18 > 0, so this critical point corresponds to a minimum.

Therefore, f has a LOCAL minimum at x=2. However, this is not an absolute minimum, because f(x) -> -infinity as x -> -infinity. There is no absolute minimum.

Answer 2

F[x]=2x^3-3x^2-12x-4
F'[x]=6x^2-6x-12=6[x^2-x-2]
=6[x+1][x-2]
If we equate it to zero we will get the
value of x for which the function can
be either max. or min.
x=-1 or 2
F"[x]=6[2x-1]
we find that the second derivative is -ve for
x=-1 and +ve for x=2
This gives us the result that for x=2
the function will have absolute min. value