Question 50: The value of m so that the function f(x) = x3 – 3×2 + 3(m2 – 1)x reaches its minimum at x0 = 2 is:

We have: y’ = 3x^{2 }– 6x + 3(m^{2} – first)

The function has a minimum at x_{0} = 2 y'(2) = 0 m = ±1

We have: y” = 6x – 6 y”(2) = 12 > 0, m

So the function has a minimum at x_{0} = 2 when m = ±1.

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