Question

6: Which of the extrema below does the the function have?

Answer 1

`f(x)=x^5-4x^3+x-1`

To find the extremas of f(x), first we must find the roots of its derivate:

`\begin{gathered} f^(\prime)(x)=5x^4-12x^2+1 \\ y=x^2 \\ f(y)=5y^2-12y+1 \\ y=(-(-12)\pm√((-12)^2-4\cdot5\cdot1))/(2\cdot5) \\ y=(12\pm2√(31))/(10) \\ y=x^2=(6\pm√(31))/(5) \\ x_1=\sqrt{(6+√(31))/(5)}\approx1.52 \\ x_2=\sqrt{(6-√(31))/(5)}\approx0.29 \\ x_3=-\sqrt{(6-√(31))/(5)}\approx-0.29 \\ x_4=-\sqrt{(6+√(31))/(5)}\approx-1.52 \end{gathered}`

Now, we must identify the signal of f'(x) for every interval in order to check if it has local maximuns of minimuns:

`\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}`

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.