Find the extreme values for the function:f(x) = 3x^2 - 5x + 3 Solution f(x) = 3x^2 - 5x + 3 First we will differentiate . f\'(x) = 6x - 5 We will find the deicative\'s zeros. ==> 6x - 5 = 0 ==> x= 5/6 Then the function has an extreme value when x= 5/6 ==> f(5/6) = 3*(5/6)^2 - 5(5/6) +3 = 3*25/36 - 25/6 + 3 = (75 - 150 + 108)/3 = 33/3 = 11 Since the factor for x^2 is positive, then the function has MINIMUM value at f(5/6) = 11.