use technology to graph the function y = x³ + 2x² - x - 2 in the window -3, 3 × -4, 4. then find each of the…

Answer:

1. Local - maximum value:
   • First, find the derivative of the function (y = x^{3}+2x^{2}-x - 2). Using the power - rule ((x^{n})^\prime=nx^{n - 1}), we have (y^\prime=3x^{2}+4x - 1).
   • Set (y^\prime = 0) to find the critical points. Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (3x^{2}+4x - 1 = 0), where (a = 3), (b = 4), and (c=-1).
     • (x=\frac{-4\pm\sqrt{4^{2}-4\times3\times(-1)}}{2\times3}=\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}\approx\frac{-2 + 2.65}{3}\approx0.22) and (x_2=\frac{-2-\sqrt{7}}{3}\approx\frac{-2-2.65}{3}\approx - 1.55).
   • Then, use the second - derivative test. Find the second - derivative (y^{\prime\prime}=6x + 4).
     • For (x=\frac{-2+\sqrt{7}}{3}), (y^{\prime\prime}=6\times\frac{-2+\sqrt{7}}{3}+4=-4 + 2\sqrt{7}+4=2\sqrt{7}\approx5.29>0), so this is a local minimum.
     • For (x=\frac{-2-\sqrt{7}}{3}), (y^{\prime\prime}=6\times\frac{-2-\sqrt{7}}{3}+4=-4-2\sqrt{7}+4=-2\sqrt{7}\approx - 5.29<0), so this is a local maximum.
     • Substitute (x=\frac{-2-\sqrt{7}}{3}) into the original function (y = x^{3}+2x^{2}-x - 2).
       • (y=(\frac{-2-\sqrt{7}}{3})^{3}+2(\frac{-2-\sqrt{7}}{3})^{2}-(\frac{-2-\sqrt{7}}{3})-2).
       • First, ((\frac{-2-\sqrt{7}}{3})^{2}=\frac{4 + 4\sqrt{7}+7}{9}=\frac{11 + 4\sqrt{7}}{9}), and ((\frac{-2-\sqrt{7}}{3})^{3}=\frac{(-2-\sqrt{7})(11 + 4\sqrt{7})}{27}=\frac{-22-8\sqrt{7}-11\sqrt{7}-28}{27}=\frac{-50 - 19\sqrt{7}}{27}).
       • (y=\frac{-50 - 19\sqrt{7}}{27}+2\times\frac{11 + 4\sqrt{7}}{9}+\frac{2+\sqrt{7}}{3}-2).
       • (y=\frac{-50 - 19\sqrt{7}+2(33 + 12\sqrt{7})+9(2+\sqrt{7})-54}{27}).
       • (y=\frac{-50 - 19\sqrt{7}+66 + 24\sqrt{7}+18 + 9\sqrt{7}-54}{27}=\frac{-50+66 + 18-54+( - 19 + 24+9)\sqrt{7}}{27}=\frac{-10 + 14\sqrt{7}}{27}\approx0.96).
   • The local maximum value is approximately (0.96) and it occurs at (x\approx - 1.55).
2. Local - minimum value:
   • Since (x=\frac{-2+\sqrt{7}}{3}\approx0.22) is a local minimum (from the second - derivative test above).
   • Substitute (x = \frac{-2+\sqrt{7}}{3}) into the original function (y = x^{3}+2x^{2}-x - 2).
     • (y=(\frac{-2+\sqrt{7}}{3})^{3}+2(\frac{-2+\sqrt{7}}{3})^{2}-(\frac{-2+\sqrt{7}}{3})-2).
     • After calculation, (y\approx - 2.04). The local minimum value is approximately (-2.04) and it occurs at (x\approx0.22).
3. Interval of decrease:
   • The function (y) is decreasing when (y^\prime=3x^{2}+4x - 1<0).
   • The roots of (y^\prime = 0) are (x=\frac{-2\pm\sqrt{7}}{3}).
   • The solution to (3x^{2}+4x - 1<0) is the interval ((\frac{-2-\sqrt{7}}{3},\frac{-2+\sqrt{7}}{3})), which is approximately ((-1.55,0.22)) in interval notation.

Explanation:

Step1: Find the derivative

Differentiate (y = x^{3}+2x^{2}-x - 2) to get (y^\prime=3x^{2}+4x - 1).

Step2: Find critical points

Set (y^\prime = 0) and use quadratic formula (x=\frac{-4\pm\sqrt{4^{2}-4\times3\times(-1)}}{6}=\frac{-2\pm\sqrt{7}}{3}).

Step3: Use second - derivative test

Find (y^{\prime\prime}=6x + 4). Evaluate at critical points to determine max/min.

Step4: Find local max/min values

Substitute critical points into original function.

Step5: Find interval of decrease

Solve (y^\prime<0) using the critical points.