91Ó°ÊÓ

We have a degree of five from $f\left(x\right)=4x^5-8x^4-x+2$. Thus, the graph of function has an end behavior similar to$y=4x^5$

Determine the intercept of function to get the x- intercept.

$\begin{array}{r}4x^5−8x^4−x+2â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =0\\ 4x^4(x−2)−(x−2)â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =0\\ \left(4x^4−1\right)(x−2)â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =0\\ 2x^2+1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =0;2x^2−1=0x−2=0\\ 2x^2â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =−1;2x^2=1;x=2\\ \sqrt{x^2}â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =\sqrt{−\frac{1}{2};}\sqrt{x^2}=\sqrt{\frac{1}{2}}\\ xâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =\frac{\sqrt{2}}{2};x=-\frac{\sqrt{2}}{2};\end{array}$

Considering$f\left(0\right)=2$, thus y-intercept is$(0,2)$

From the graph above, we can see that the maximum and minimum turning point of the function is:

$(-0.298,2.226)$and$(1.612,-10.092)$

Therefore, the domain and range of the function is the set of all real numbers.

By observing the graph above, we say that the function at the intervals of $(-∞,-0.298)$and $(1.612,∞)$is increasing.

At intervals of$(-0.298,1.612)$decreasing.