Question

The number of critical points for $(x+3){(3x-2)}^5{(7-x)}^3{(5x+8)}^2\ge 0$ is

Answer 1

We have $(x+3){(3x-2)}^5{(7-x)}^3{(5x+8)}^2\ge 0$
$-(x+3){(3x-2)}^5{(x-7)}^3{(5x+8)}^2\ge 0$
$\Rightarrow (x+3){(x-\frac{2}{3})}^5{(x-7)}^3{(x+\frac{8}{5})}^2\le 0$
$\therefore$ The critical points are
-3,$-\frac{8}{5}$, $\frac{2}{3}$ , 7
$\therefore x\in (-\infty ,-3)\cup [\frac{2}{2},7]\cup \{-\frac{8}{5}\}$.