SOLUTION: If x is a real number, find the minimum value of 3^(x²-4x+8)

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Question 1199922: If x is a real number, find the minimum value of 3^(x²-4x+8)
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
The function will be a minimum at the point where the exponent is a minimum.
Thus d/dx(x^2-4x+8) = 0 = 2x - 4 -> x = 2