Answer to Question #26689 in Algebra for Jackie
2x^2 -12x + 19
F(x) = 2x² -12x + 19
Let's find the derivative of F(x) with respect to x:
F'(x) = 4x - 12
Now let's find zeros of the derivative:
F'(x) = 0 ==> x = 3
The intervals of sign constancy are
F'(x)<0, if x<3
F'(x)>0, if x>3
Therefore, F is decreasing on the interval (-∞,3) and increasing on the interval (3,+∞).
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