Find the intervals on which $f$ is increasing or decreasing. $f(x)= x^{3}-3x+1$

Question

Note: This is part 2nd of a 6 part question, split as 6 separate questions here.

Answer 1

Toolbox:

• (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
• (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$

$f(x)=x^3-3x+1$

Step 1:

$f'(x)=3x^2-3=3(x^2-1)=3(x+1)(x-1)$

$f'(x)=0\;when \;x=-1,1$

Step 2:

Interval $-\infty < x < -1$

$(x+1)\;= -$

$(x-1)\;=-$

$f'c (x)\;=+$

Interval of inc /dec $f(x)$= $f(x)$ is increasing in $(-\infty,-1]$

Interval $-1 < x<1$

$(x+1)\;= +$

$(x-1)\;=-$

$f'c (x)\;=-$

Interval of inc /dec $f(x)$= $f(x)$ is decreasing in $[1,-1]$

Interval $1 < x <\infty$

$(x+1)\;= +$

$(x-1)\;=+$

$f'c (x)\;=+$

Interval of inc /dec $f(x)$= $f(x)$ is increasing in $[1,\infty]$

$\therefore \;f(x)$ increasing in $(- \infty,-1] \cup [1, \infty)$ and decreasing in $[-1,1]$