
This is a graph of the equation 2X^{3} 7X^{2} 5X +4 = 0. Now we are dealing with cubic equations instead of quadratics. From Part I we know that to find minimums and maximums, we determine where the equation's derivative equals zero. The equation's derivative is 6X^{2} 14X 5 and when this derivative equals zero 6X^{2} 14X 5 = 0 the roots of the derivative are 2.648 and .3147
Notice we have 2 extreme points, whereas in Part I, we only had one. Inputting 2.648 into the second derivative we get 17.776 which is greater than zero so we have a minimum at that point. When X= .3147, the second derivative value is 17.776 and being less than zero we have a maximum. Of course if you look at the graph, you can see that the left side of the graph starts from negative infinity, and the right side goes on to infinity. So, in this case we are talking about a relative maximum at point X = .3147 and a relative minimum at point X= 2.648. To obtain the 'Y' values, we input 2.648 and .3147 into the original equation 2X^{3} 7X^{2} 5X +4 = 0 , and we get values of 21.188 and 4.818 respectively.
So, the relative minimum is at (X= 2.648, Y= 21.188) Click here to go to Part III

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