Question

Find the values of x for which the function f(x)=x^3–3x, is increasing.

Answer 1

Solution. Find the first derivative of the function and find the values when the derivative is zero.

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

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

$$x^2=1$$

The roots of the equation are

$$x_1=-1$$ $$x_2=1$$

Find the value of the derivative for each of the intervals.

For

$$x\in (-\infty, -1)$$

$$f'(x)>0$$

function f(x) is increases.

For

$$x\in (-1,1)$$

$$f'(x)<0$$

function f(x) is decreases.

For

$$x\in (1,\infty)$$

$$f'(x)>0$$

function f(x) increases. Therefore the values of x for which the function f(x)=x^3–3x, is increasing

$$x\in(-\infty,-1)\bigcup(1,\infty)$$

Answer. 

$$x\in(-\infty,-1)\bigcup(1,\infty)$$