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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:
one maximum and one minimum value
Given, $f(x)=x^3+a x^2+b x+c, a^2 \leq 3 b$.
On differentiating w.r.t. $x$, we get
$\begin{aligned} & f^{\prime}(x)=3 x^2+2 a x+b \\ & \text { Put } \quad f^{\prime}(x)=0 \\ & \Rightarrow \quad 3 x^2+2 a x+b=0 \\ & \end{aligned}$
$\begin{aligned}
\Rightarrow \quad x & =\frac{-2 a \pm \sqrt{4 a^2-12 b}}{2 \times 3} \\
& =\frac{-2 a \pm 2 \sqrt{a^2-3 b}}{3}
\end{aligned}$
Since, $\quad a^2 \leq 3 b$, $\therefore \quad x$ has an imaginary value. Hence, no extreme value of $x$ exist.
On differentiating w.r.t. $x$, we get
$\begin{aligned} & f^{\prime}(x)=3 x^2+2 a x+b \\ & \text { Put } \quad f^{\prime}(x)=0 \\ & \Rightarrow \quad 3 x^2+2 a x+b=0 \\ & \end{aligned}$
$\begin{aligned}
\Rightarrow \quad x & =\frac{-2 a \pm \sqrt{4 a^2-12 b}}{2 \times 3} \\
& =\frac{-2 a \pm 2 \sqrt{a^2-3 b}}{3}
\end{aligned}$
Since, $\quad a^2 \leq 3 b$, $\therefore \quad x$ has an imaginary value. Hence, no extreme value of $x$ exist.
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