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Prove that the following functions are not monotonic in the intervals given. $2x^{2}+x-5$ on $[-1 ,0 ]$

Note: This is part 1st of a 4 part question, split as 4 separate questions here.

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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 $I$
$2x^2+x-5\;on\; [-1,0]$
Step 1:
$f'(x)=2x+1$
$f'(x) >0=>2x+1 >0 \;or \;x > \large\frac{-1}{2}$
when $x > -\large\frac{1}{2}$ $f'(x) > 0 $ and when $x < -\large\frac{1}{2} $$f'(x) <0$
$-\large\frac{1}{2} $$\in [-1,0]$
$\therefore \;f'(x) $ takes both negative and positive values on $[-1,0]$.
So it is not monotonic in $[-1,0]$
answered Jul 30, 2013 by meena.p
 
0 votes
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 $I$
Step 1:
$f'(x)=2x+1$
When $x=-\large\frac{1}{4}$$ \in [-1,0]$
$f'(x) =-2 -\large\frac{1}{4} $$+1=-\large\frac{1}{2}$$+1=\large\frac{1}{2} $$>0$
Step 2:
When $x=-\large\frac{3}{4}$$ \in [-1,0]$
$f'(x)=-2 \times \large\frac{3}{4}$$ +1$
$\qquad=\large\frac{-3}{2}$$+1=\large\frac{-1}{2} < 0$
$f'(x) $ takes both positive and negative values in $[-1,0]$
So it is not monotonic in $[-1,0]$
answered Jul 30, 2013 by meena.p
 

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