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Which of the following function are increasing or decreasing on the interval given? \[\]$2x^{2}+3x$ on $[-\large\frac{1}{2} ,\frac{1}{2}]$

Note: This is part 2nd of a 5 part question, split as 5 separate questions here.

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 $I$
$f(x)= 2 x^2+3x$ on $\bigg[-\large\frac{1}{2}, \frac{1}{2}\bigg]$
$f'(x)=4x+3$
$f'(x) > 0\;where \; x > -\large\frac{3}{4}$
In $\bigg[-\large\frac{1}{2}, \frac{1}{2}\bigg] x > -\large\frac{3}{4}$
$ \therefore f'(x) >0$
So $f(x) $ is strictly increasing
answered Jul 30, 2013 by meena.p
 

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