f(x) = 8x^{2} − 16x + 1

Hello!

This f(x) is a quadratic function with the positive coefficient at x^2.So it has one minimum at its vertex.

Make the perfect square:f(x) = 8x^2 - 16x + 1 = 8(x^2 - 2x) + 1 = 8((x-1)^2 -1) +1 = 8(x-1)^2 - 7.

This cannot be <-7 ((x-1)^2 >= 0), and =-7 only at x=1.

Answer: the minimum value is -7. (it is reached at x=1)

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