Monotonicity of Functions: Problems with Solutions

Solution:
$f(x)=3x-3$
$f'(x)=(3x-3)'=(3x)'-3'=3-0=3$

$f'(x)>0\Rightarrow$ Function is increasing on $(-\infin, +\infin)$

Solution:
$f(x)=x^2-2x-3$

$f'(x)=(x^2-2x-3)'=(x^2)'-(2x)'-(3)'=2x-2-0=2x-2$

$f'(x)=0$
$2x-2=0$
$x=1$

$f(x)$ is decreasing on $(-\infin,1)$
$f(x)$ is increasing on $(1,+\infin)$

Solution:
$f(x)=\sqrt{1-x}$

Domain: $1-x \ge 0 \Leftrightarrow x \le 1$

$f'(x)=(\sqrt{1-x})'=((1-x)^{\frac{1}{2}})'=\frac{1}{2\sqrt{1-x}}(1-x)'=\frac{1}{2\sqrt{1-x}}(-1)=-\frac{1}{2\sqrt{1-x}}<0$

The function is decreasing for every value on the domain.

The function is decreasing on $(-\infin,1]$

Solution:
$f'(x)=3x^2-12x+9$

$f'(x)=0$
$3x^2-12x+9=0$
$x^2-4x+3=0$
$D=16-12=4$
$x_1=3$
$x_2=1$

$f(x)$ increases on $(-\infin,1]\cup[3,\infin)$
$f(x)$ decreases on $[1, 3]$

Solution:
$f'(x)=(-x^3+6x^2-9x)'=(-x^3)'+(6x^2)'-(9x)'=-3x^2+12x-9$

$f'(x)=0$
$-3x^2+12x-9 = 0 \ \ \ \ \div(-3)$
$x^2-4x+3 = 0$
$D=16-12=4$
$x_1=3$
$x_2=1$

The function is increasing on $(-\infin,1]\cup[3,+\infin)$ and decreasing on $[1,3]$