Question

10. Determine the open interval(s) on which the function f(x) = 12x - x^3 is increasing.

Answer 1

Answer:

$\large\boxed{x\in(-2,\ 2)}$

Step-by-step explanation:

$f(x)=12x-x^3$

Calculate the derivative of the function:

$f'(x)=\left(12x-x^3\right)'=(12x)'-\left(x^3\right)'=12-3x^2$

Calculate the zeros of the derivative:

$f'(x)=0\Rightarrow12-3x^2=0\qquad\text{add}\ 3x^2\ \text{to both sides}\\\\3x^2=12\qquad\text{divide both sides by 3}\\\\x^2=4\to x=\pm\sqrt4\\\\x=\pm2$

Let's check the derivative sign:

$f'(x)>0\iff12-3x^2>0\qquad\text{use the distributive property}\\\\-3(x^2-4)>0\\\\-3(x^2-2^2)>0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\-3(x-2)(x+2)>0\\^{\qquad x=2\qquad x=-2$

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:

$x\in(-2,\ 2)$