The First and Second Derivatives - Dartmouth College

The critical points of g(x) are precisely the values of x where the derivative of g(x) is 0, so we set the formula above equal to 0 and solve the resulting quadratic equation: 3x2 ¡18x+15 = 0 x2 ¡6x+5 = 0 (x¡1)(x¡5) = 0 x = 1 or x = 5: So the critical points of g(x) are x = 1 and x = 5. We now want to apply the second derivative test, and to ................
................