The interval of $f (x) = x^{3} + 2 x^{2} + 5 x, x < 0$ for which it is concave upwards is

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Solution

$f (x) = x^{3} + 2 x^{2} + 5 x, \forall x < 0 \dots (i)$
$f^{'} (x) = 3 x^{2} + 4 x + 5$
$\Rightarrow f^{''} (x) = 6 x + 4$
For $f$ to be concave upward,
$f^{''} (x) > 0$
$\Rightarrow x > - \frac{2}{3} \dots (i i)$
From $(i) and (i i),$ we get
Interval of concave upwards is $(- \frac{2}{3}, 0)$

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