Answer: -11
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Work Shown:
t = x³
f(x) = 1 + 12x³ + 3x⁶
f(x) = 1 + 12x³ + 3(x³)²
f(t) = 1 + 12t + 3t²
f(t) = 3t² + 12t + 1
f(t) = 3(t² + 4t) + 1
f(t) = 3(t² + 4t + 4 - 4) + 1 .... see note below
f(t) = 3( (t²+4t+4) - 4) + 1
f(t) = 3( (t+2)² - 4) + 1
f(t) = 3(t+2)² - 3*4 + 1
f(t) = 3(t+2)² - 12 + 1
f(t) = 3(t+2)² - 11
The smallest that t² can get is 0. By extension, the smallest (t+2)² can get is also 0, and same goes for 3(t+2)².
So effectively we replace the 3(t+2)² portion with 0 and end up with 3(t+2)²-11 = 0-11 = -11 as the minimum value of f(x)
This is the smallest output possible. Confirmation of such is shown in the graph below.
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Note: Recall that (A+B)² = A²+2AB+B². If we use this rule on (t+2)², then we get t²+4t+4. Earlier we simply had t²+4t without the extra +4 at the end. So I added 4 and subtracted 4 to balance things out. This completed the square.