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

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

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  • (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 $[-1,1]$
answered Jul 30, 2013 by meena.p
 

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