use technology to graph the function ( y = x^{3}+2x^{2}-x - 2 ) in the window ( -3,3\times-4,4 ). then find…

Explanation:

Step1: Find the derivative of the function

The function is (y = x^{3}+2x^{2}-x - 2). The derivative (y'=3x^{2}+4x - 1).

Step2: Find the critical points

Set (y' = 0), so (3x^{2}+4x - 1=0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) where (a = 3), (b = 4), (c=-1). [x=\frac{-4\pm\sqrt{16+12}}{6}=\frac{-4\pm\sqrt{28}}{6}=\frac{-4\pm2\sqrt{7}}{6}=\frac{-2\pm\sqrt{7}}{3}] (x_1=\frac{-2+\sqrt{7}}{3}\approx0.22), (x_2=\frac{-2 - \sqrt{7}}{3}\approx - 1.55)

Step3: Use the second - derivative test

The second - derivative (y'' = 6x+4). For (x=\frac{-2+\sqrt{7}}{3}), (y''=6\times\frac{-2+\sqrt{7}}{3}+4=2(-2+\sqrt{7})+4 = 2\sqrt{7}\approx5.29>0), so (x=\frac{-2+\sqrt{7}}{3}) is a local minimum. For (x=\frac{-2-\sqrt{7}}{3}), (y''=6\times\frac{-2-\sqrt{7}}{3}+4=2(-2-\sqrt{7})+4=-2\sqrt{7}\approx - 5.29<0), so (x=\frac{-2-\sqrt{7}}{3}) is a local maximum. Substitute (x = \frac{-2-\sqrt{7}}{3}) into the original function (y=x^{3}+2x^{2}-x - 2) [y=\left(\frac{-2-\sqrt{7}}{3}\right)^{3}+2\left(\frac{-2-\sqrt{7}}{3}\right)^{2}-\left(\frac{-2-\sqrt{7}}{3}\right)-2] [=\frac{(-2-\sqrt{7})^{3}+2(-2 - \sqrt{7})^{2}+3(2+\sqrt{7})-18}{9}] [=\frac{(-8-12\sqrt{7}-21 - 7\sqrt{7})+2(4 + 4\sqrt{7}+7)+6 + 3\sqrt{7}-18}{9}] [=\frac{(-29-19\sqrt{7})+(22 + 8\sqrt{7})+6 + 3\sqrt{7}-18}{9}] [=\frac{(-29 + 22+6-18)+(-19\sqrt{7}+8\sqrt{7}+3\sqrt{7})}{9}=\frac{-19-8\sqrt{7}}{9}\approx0.39)

Answer:

(0.39)