absolute min?

Question

find the absolute min at: f(x) = x^6(x-2)^7 on interval (-12,14)

Answer 1

absolute minimum occurs when x = -12
min = -3.147630363 X 10^14

Answer 2

f(x) is a polynomial of degree 13. To find the absolute minimum, find f(-12), f(14) and local minimum and select the absolute minimum from them all.
f(-12) = (-12)^6 * (-12 - 2)^7 = - 314763036327936
f(14) = (14)^6 * (14 -2)^7 = 269796888281088

f'(x) = 6x^5 * (x -2)^7 + 7x^6 * (x - 2)^6
f'(x) = x^5 * (x - 2)^6 * [ 6x - 12 + 7x ]
f'(x) = 0 => x = 0 or x = 2 or x = 12/13
None of these values are between -12 and 14.
Between absolute values of f(-12) and f(14),
f(14) = 269796888281088 is minimum.

If the smallest value is meant, then
f(-12) = - 314763036327936 being negative is the least.

Answer 3

f'(x) = 6x^5(x-2)^7 + 7x^6*(x-2)^6
f' (x) =x^5*(x-2)^6* [ 6(x-2)+7x]
f ' (x) = x^5*(x-2)^6*(13x-12)

if x< 0 then f '(x) >0
if 0 < x < 12/13 then f ' (x) <0
and if x> 12/13 then f '(x) >0

the absolute min can be for x = -12 or for x= 12/13

the calculus is easy f(12/13) > f(-12)
so the absolute min is f( -12)

Answer 4

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