Answer:
[tex]\large\boxed{x\in(-2,\ 2)}[/tex]
Step-by-step explanation:
[tex]f(x)=12x-x^3[/tex]
Calculate the derivative of the function:
[tex]f'(x)=\left(12x-x^3\right)'=(12x)'-\left(x^3\right)'=12-3x^2[/tex]
Calculate the zeros of the derivative:
[tex]f'(x)=0\Rightarrow12-3x^2=0\qquad\text{add}\ 3x^2\ \text{to both sides}\\\\3x^2=12\qquad\text{divide both sides by 3}\\\\x^2=4\to x=\pm\sqrt4\\\\x=\pm2[/tex]
Let's check the derivative sign:
[tex]f'(x)>0\iff12-3x^2>0\qquad\text{use the distributive property}\\\\-3(x^2-4)>0\\\\-3(x^2-2^2)>0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\-3(x-2)(x+2)>0\\^{\qquad x=2\qquad x=-2[/tex]
Sketch a parabola open down
(before the brackets there is the number -3 < 0)
Look at the picture.
Where the derivative is negative, the function decreases.
Where the derivative is positive, the function increases.
Therefore your answer is:
[tex]x\in(-2,\ 2)[/tex]
