f(x)=3x^2-6x+6 where does the minimum or maximum value occur? What is the functions minimum or maximum value?

Answer 1

3x^2 -6x + 6 = 0

divide by 3

x^2 -2x + 2 = 0

complete the square

take half the coefficient of the x term and square it

that is 1. Add and subtract 1

x^2 -2x + 1 + 2 -1 = 0

(x-1)^2 +1 = 0

It's a parabola with vertex (1,1)

the vertex is the minimum point for a parabola opening upwards

the maximum value is +infinity

or rather it approaches infinity, but never reaches a "maximum"

another way to solve this is take the derivative and set it = 0

6x - 6 = 0

6x = 6

x=6/6 = 1

y=3(1)^2 - 6(1) + 6 = 1

the point (1,1) is the minimum of the function