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# Prove that the following functions are not monotonic in the intervals given. $x(x-1)(x+1)$ on $[0 , 2 ]$

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## 1 Answer

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Toolbox:
• If $f(x)$ is defined on an open interval $I$ and $f'(x)$ takes both positive and negative values on $I$, then $f$ is not monotonic on $I$
$f(x) =x (x-1)(x+1) \;on\; [0,2]$
Step 1:
$f(x)=x(x^2-1)$
$\qquad=x^3-x$
Step 2:
$f'(x)=x^2-1$
=>$f'(x) =(x+1)(x-1)$
$f '(x)< 0$ when $x \in (-1,1)$ and $f'(x) >0$ when x axes outside $[-1,1]$
Step 3:
Now, $1 \in [0,2]$, when $0 < x < 1, f'(x) > 0$
and when $1 < x <2 , f'(x) <0$
$\therefore$ the function changes sign in $[-1,1]$
It is not monotonic.

answered Jul 30, 2013 by

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