## Calculus: Early Transcendentals 8th Edition

Original Equation: ${f(x) = 2x^{3} -x^{2} +2x}$ The critical numbers of a function are found when the derivative of a function is set to zero. Finding the derivative of the function using the power rule: ${f'(x) = 6x^{2} - 2x+2}$ Setting the derivative to zero and factoring out the 2: ${f'(x) = 2(3x^{2} - x+1) = 0}$ There is no way to solve this equation; plugging the values into the quadratic equation yields complex roots. This means there are no critical numbers. To prove that there are 0 real roots, we can find the discriminant of the equation as given by the formula ${b^{2}-4ac}$. Plugging in values gives ${1^{2} - 4(3)(1) = 1-12 = -11}$ -11${\lt}$0 therefore there are no real roots meaning there are no critical numbers.