[tex]f(x)=x^5-4x^3+x-1[/tex]
To find the extremas of f(x), first we must find the roots of its derivate:
[tex]\begin{gathered} f^{\prime}(x)=5x^4-12x^2+1 \\ y=x^2 \\ f(y)=5y^2-12y+1 \\ y=\frac{-(-12)\pm\sqrt{(-12)^2-4\cdot5\cdot1}}{2\cdot5} \\ y=\frac{12\pm2\sqrt{31}}{10} \\ y=x^2=\frac{6\pm\sqrt{31}}{5} \\ x_1=\sqrt{\frac{6+\sqrt{31}}{5}}\approx1.52 \\ x_2=\sqrt{\frac{6-\sqrt{31}}{5}}\approx0.29 \\ x_3=-\sqrt{\frac{6-\sqrt{31}}{5}}\approx-0.29 \\ x_4=-\sqrt{\frac{6+\sqrt{31}}{5}}\approx-1.52 \end{gathered}[/tex]
Now, we must identify the signal of f'(x) for every interval in order to check if it has local maximuns of minimuns:
[tex]\begin{gathered} f^{\prime}(x)>0,\text{ }if\text{ }xx_3 \\ f^{\prime}(x)>0,ifx_3\lt x\gt x_2 \\ f^{\prime}(x)\lt0,ifx_2\lt x\gt x_1 \\ f^{\prime}(x)\gt0,ifx>x_1 \end{gathered}[/tex]
Therefore, the function have local maximuns and minumuns. But, since it goes to negative infinity, if x goes to negative infinity, or to positive infinity, if x goes to positive infinity, it hasn't a global maximum or minimum.
Therefore, only statements III and IV are correct.
Answer: D.
