Mohon bantuannya pakai jalan ya.

2)
turun berarti f'(x) < 0
[tex]2. \\ f(x) = {x}^{3} + 3 {x}^{2} - 9x + 5 \\ \\ {f}^{.} (x) = 3 {x}^{2} + 6x - 9 \\ 0 > {x}^{2} + 2x - 3 \\ (x - 1)(x + 3) \\ x = 1 \: dan \: x = - 3 [/tex]
untuk gambarnya, maaf nggak bisa nyertain.
-3 < x < 1

[tex]3) \\ f(x) = 3 {x}^{4} + 4 {x}^{3} - 12 {x}^{2} + 5 \\ f ^{.} (x) = 12 {x}^{3} + 12 {x}^{2} - 24x \\ 0 < 12 {x}^{3} + 12 {x}^{2} - 24x \\ 0 < {x}^{3} + {x}^{2} - 2x \\ 0 < x( {x}^{2} + x - 2) \\ 0 < x(x - 1)(x + 2)[/tex]
x>0 dan -2<x<1