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Find the intervals on which $f$ is increasing or decreasing. $f(x)=20-x-x^{2}$

Note: This is part 1st of a 6 part question, split as 6 separate questions here.

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1 Answer

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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 $I$
$f(x)= 20-x-x^2$
Step 1:
$f'(x) =-1-2x$
$f'(x)=0=>-1-2x=0=>x =\large\frac{-1}{2}$
Step 2:
Interval $(-\infty, -\large\frac{1}{2}]$
$f'(x)=+$
$f(x)$ inc/dec in interval=$f(x)$ is increasing in $(-\infty, \large\frac{1}{2}]$
Interval $[\large\frac{1}{2}$$,\infty)$
$f'(x)=-$
$f(x)$ inc/dec in interval=$f(x)$ is decreasing in $[\large\frac{1}{2},\infty)$
answered Jul 30, 2013 by meena.p
 

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