Question

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

Answer 1

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