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The function $f(x)=x^{3}+a x^{2}+b x+c, a^{2} \leq 3 b$ has
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Verified Answer
The correct answer is:
no extreme value
Given, $f(x)=x^{3}+a x^{2}+b x+c, a^{2} \leq 3 b$
$$
\Rightarrow f^{\prime}(x)=3 x^{2}+2 a x+b
$$
Put $f^{\prime}(x)=0$
$$
\begin{array}{l}
\Rightarrow 3 x^{2}+2 a x+b=0 \\
\Rightarrow x=\frac{-2 a \pm \sqrt{4 a^{2}-12 b}}{2 \times 3} \\
=\frac{-2 a \pm 2 \sqrt{a^{2}-3 b}}{3}
\end{array}
$$
Since, $a^{2} \leq 3 b$,
$\therefore x$ has an imaginary value. Hence, no extreme value of $x$ exists.
$$
\Rightarrow f^{\prime}(x)=3 x^{2}+2 a x+b
$$
Put $f^{\prime}(x)=0$
$$
\begin{array}{l}
\Rightarrow 3 x^{2}+2 a x+b=0 \\
\Rightarrow x=\frac{-2 a \pm \sqrt{4 a^{2}-12 b}}{2 \times 3} \\
=\frac{-2 a \pm 2 \sqrt{a^{2}-3 b}}{3}
\end{array}
$$
Since, $a^{2} \leq 3 b$,
$\therefore x$ has an imaginary value. Hence, no extreme value of $x$ exists.
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