the polynomial function f is defined by $f(x)=-3x^{3}+3x^{2}+7x - 3$. use the aleks graphing calculator to…

Explanation:

Step1: Find the derivative of the function

The function is (f(x)=-3x^{3}+3x^{2}+7x - 3). Using the power rule ((x^{n})^\prime=nx^{n - 1}), the derivative (f^\prime(x)=-9x^{2}+6x + 7).

Step2: Find the critical points

Set (f^\prime(x) = 0), so (-9x^{2}+6x + 7=0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0) (here (a=-9), (b = 6), (c = 7)). (x=\frac{-6\pm\sqrt{6^{2}-4\times(-9)\times7}}{2\times(-9)}=\frac{-6\pm\sqrt{36 + 252}}{-18}=\frac{-6\pm\sqrt{288}}{-18}=\frac{-6\pm12\sqrt{2}}{-18}=\frac{1\pm2\sqrt{2}}{3}). (x_1=\frac{1 + 2\sqrt{2}}{3}\approx1.24), (x_2=\frac{1-2\sqrt{2}}{3}\approx - 0.57).

Step3: Use the second - derivative test

Find the second - derivative (f^{\prime\prime}(x)=-18x + 6). For (x=\frac{1-2\sqrt{2}}{3}): (f^{\prime\prime}(\frac{1-2\sqrt{2}}{3})=-18\times\frac{1 - 2\sqrt{2}}{3}+6=-6 + 12\sqrt{2}+6=12\sqrt{2}>0) (so this is a local minimum). For (x=\frac{1 + 2\sqrt{2}}{3}): (f^{\prime\prime}(\frac{1 + 2\sqrt{2}}{3})=-18\times\frac{1+2\sqrt{2}}{3}+6=-6-12\sqrt{2}+6=-12\sqrt{2}<0) (so this is a local maximum).

Step4: Find the value of the function at the local maximum point

Substitute (x=\frac{1 + 2\sqrt{2}}{3}) into (f(x)): (f(\frac{1 + 2\sqrt{2}}{3})=-3(\frac{1 + 2\sqrt{2}}{3})^{3}+3(\frac{1 + 2\sqrt{2}}{3})^{2}+7(\frac{1 + 2\sqrt{2}}{3})-3). Using a calculator: (x=\frac{1 + 2\sqrt{2}}{3}\approx1.24), (f(x)\approx-3\times(1.24)^{3}+3\times(1.24)^{2}+7\times(1.24)-3) (f(x)\approx-3\times1.907+3\times1.538+8.68-3) (f(x)\approx-5.721 + 4.614+8.68-3) (f(x)\approx4.57)

Answer:

((1.24,4.57))