find the extreme values for f(x) = 3x^2 - 5x + 3 Solution f(x) = 3x^2 - 5x + 3 To find extreme values first we will find f\'(x). f\'(x) = 6x - 5 Now we find critical values which are the derivative\'s zeros. ==> 6x - 5 = 0 ==> x = 5/6 Then f(x) has an extreme values of f(5/6); f(5/6) = 3(5/6)^2 - 5(5/6) + 3 = 3*25/36 - 25/6 + 3 = (75 - 150 + 108)/36 = 33/36 = 11/6 Since the factor if x^2 is positive , then the function has MINIMUM value at f(5/6) = 11/6.