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The function $\mathrm{f}(x)=x^3-6 x^2+9 x+2$ has maximum value when $x$ is
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The correct answer is:
1
$\begin{aligned}
& \mathrm{f}(x)=x^3-6 x^2+9 x+2 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2-12 x+9
\end{aligned}$
For maximum or minimum,
$\begin{aligned}
& \mathrm{f}^{\prime}(x)=0 \\
& \Rightarrow 3 x^2-12 x+9=0 \\
& \Rightarrow 3(x-1)(x-3)=0 \\
& \Rightarrow x=1,3
\end{aligned}$
Now, $\mathrm{f}^{\prime \prime}(x)=6 x-12$
$\therefore \quad \mathrm{f}^{\prime \prime}(1)=-6 < 0$
$\therefore \quad \mathrm{f}(x)$ is maximum at $x=1$.
& \mathrm{f}(x)=x^3-6 x^2+9 x+2 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2-12 x+9
\end{aligned}$
For maximum or minimum,
$\begin{aligned}
& \mathrm{f}^{\prime}(x)=0 \\
& \Rightarrow 3 x^2-12 x+9=0 \\
& \Rightarrow 3(x-1)(x-3)=0 \\
& \Rightarrow x=1,3
\end{aligned}$
Now, $\mathrm{f}^{\prime \prime}(x)=6 x-12$
$\therefore \quad \mathrm{f}^{\prime \prime}(1)=-6 < 0$
$\therefore \quad \mathrm{f}(x)$ is maximum at $x=1$.
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