find the relative maximas and minimas of each polynomial. (f(x)=-x^{4}+3x^{2}-2x + 1). round each answer to…

Explanation:

Step1: Find the derivative of the function

Given (f(x)=-x^{4}+3x^{2}-2x + 1), using the power - rule ((x^n)^\prime=nx^{n - 1}), we have (f^\prime(x)=-4x^{3}+6x - 2).

Step2: Set the derivative equal to zero

(-4x^{3}+6x - 2 = 0). Factor out (-2) to get (2x^{3}-3x + 1=0). By trial - and - error, we find that (x = 1) is a root. Then we can factor (2x^{3}-3x + 1=(x - 1)(2x^{2}+2x - 1)). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (2x^{2}+2x - 1) where (a = 2), (b = 2), (c=-1), we get (x=\frac{-2\pm\sqrt{4+8}}{4}=\frac{-2\pm2\sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}). So the critical points are (x = 1), (x=\frac{-1+\sqrt{3}}{2}\approx0.37), (x=\frac{-1 - \sqrt{3}}{2}\approx - 1.37).

Step3: Use the second - derivative test

Find the second - derivative (f^{\prime\prime}(x)=-12x^{2}+6). For (x=\frac{-1+\sqrt{3}}{2}), (f^{\prime\prime}(\frac{-1+\sqrt{3}}{2})=-12(\frac{-1 + \sqrt{3}}{2})^{2}+6=-12(\frac{1-2\sqrt{3}+3}{4})+6=-3(4 - 2\sqrt{3})+6=-12 + 6\sqrt{3}+6=6\sqrt{3}-6\approx4.39>0), so (f(x)) has a local minimum at (x=\frac{-1+\sqrt{3}}{2}), and (f(\frac{-1+\sqrt{3}}{2})=-(\frac{-1+\sqrt{3}}{2})^{4}+3(\frac{-1+\sqrt{3}}{2})^{2}-2(\frac{-1+\sqrt{3}}{2})+1\approx0.2). For (x=\frac{-1 - \sqrt{3}}{2}), (f^{\prime\prime}(\frac{-1 - \sqrt{3}}{2})=-12(\frac{-1 - \sqrt{3}}{2})^{2}+6=-12(\frac{1 + 2\sqrt{3}+3}{4})+6=-3(4 + 2\sqrt{3})+6=-12-6\sqrt{3}+6=-6 - 6\sqrt{3}\approx - 16.39<0), so (f(x)) has a local maximum at (x=\frac{-1 - \sqrt{3}}{2}), and (f(\frac{-1 - \sqrt{3}}{2})=-(\frac{-1 - \sqrt{3}}{2})^{4}+3(\frac{-1 - \sqrt{3}}{2})^{2}-2(\frac{-1 - \sqrt{3}}{2})+1\approx4.8). For (x = 1), (f^{\prime\prime}(1)=-12\times1^{2}+6=-6<0), so (f(x)) has a local maximum at (x = 1), and (f(1)=-1^{4}+3\times1^{2}-2\times1 + 1=1).

Answer:

Maxima: (-1.4,4.8); Maxima: (1,1); Minima: (0.4,0.2)