1. You did it correctly.
if the first term (x-1)^2 is positive for all x, then that term +6 is positive for all x.
2. Your f' does not match the degree of f(x). I don't know what you meant.
Please help! These 2 questions are very similar, and i tried to do completing square method, but i cant get the correct answer! I'm trying to get the dy/dx to be a complete square thingy to show that the value is always greater or equal to 0.
1. Show that the function f(x)= x^3 - 3x^2 + 9x - 5 is increasing with x for all real values of x.
f'(x)=3x^2-6x+9
=3(x^2-2x+3)
=3[x^2 - 2x + (-2/2)^2 + 3 - (-2/2)^2]
= 3[ (x-1)^2 + 2]
=3(x-1)^2 + 6
?!! The correct answer is supposed to be f'(x)=3(x-1)^2 is ≥0 for all real values of x. Did i do the completing square wrongly?? This same thing happened with the second question
2. Show that f(x) = -2x^2 + 3x^2 - 2x + 4 decreases for all real values of x.
The correct answer for this one is f'(x) = -6(x-1/2)^2 - 1/2 <0 for all real values of x
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