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If $f(x)=x^{3}+b x^{2}+c x+d$ and $0 < b^{2} < c$, then in $(-\infty, \infty)$
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The correct answer is:
$f(x)$ is a strictly increasing function
$f(x)=x^{3}+b x^{2}+c x+d, 0 < b^{2} < c$ $\therefore \quad f^{\prime}(x)=3 x^{2}+2 b x+c$ Discriminant $=4 b^{2}-12 c=4\left(b^{2}-3 c\right) < 0$ $\therefore f^{\prime}(x)>0 \forall x \in R$
Thus, $f(x)$ is strictly increasing $\forall x \in R$
Thus, $f(x)$ is strictly increasing $\forall x \in R$
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