If $\mathrm{f}(x)=x^3+\mathrm{b} x^2+\mathrm{c} x+\mathrm{d}$ and $0 \lt \mathrm{b}^2 \lt \mathrm{c}$, then in $(-\infty, \infty)$

Question

Aicharya · Accepted Answer

\begin{aligned} & f(x)=x^3+b x^2+c x+d \\ & f^{\prime}(x)=3 x^2+2 b x+c \end{aligned} $\therefore$
Now its discriminant $=4\left(b^2-3 c\right)$ $\Rightarrow 4\left(b^2-\mathrm{c}\right)-8 \mathrm{c} \lt 0$, as $\mathrm{b}^2 \lt \mathrm{c}$ and $\mathrm{c}\gt0$ $\Rightarrow \mathrm{f}^{\prime}(x)\gt0$ for all $x \in \mathrm{R}$ $\Rightarrow f$ is strictly increasing on $R$.

If $\mathrm{f}(x)=x^3+\mathrm{b} x^2+\mathrm{c} x+\mathrm{d}$ and $0 \lt \mathrm{b}^2 \lt \mathrm{c}$, then…

Solution