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If $f:[2,3] \rightarrow \mathrm{R}$ is defined by $f(x)=x^{3}+3 x-2$, then the range $f(x)$ is contained in the interval
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The correct answer is:
$[12,34]$
\begin{array}{l}
\text { Given, } f(x)=x^{3}+3 x-2 \\
\Rightarrow f^{\prime}(x)=3 x^{2}+3 \\
\text { Put } f^{\prime}(x)=0 \Rightarrow 3 x^{2}+3=0 \\
\Rightarrow x^{2}=-1
\end{array}
$\Rightarrow x^{2}=-1$
$\therefore f(x)$ is either increasing or decreasing At $x=2, f(2)=2^{3}+3(2)-2=12$ At $x=3, f(3)=3^{3}+3(3)-2=34$ $\therefore f(x) \in[12,34]$
\text { Given, } f(x)=x^{3}+3 x-2 \\
\Rightarrow f^{\prime}(x)=3 x^{2}+3 \\
\text { Put } f^{\prime}(x)=0 \Rightarrow 3 x^{2}+3=0 \\
\Rightarrow x^{2}=-1
\end{array}
$\Rightarrow x^{2}=-1$
$\therefore f(x)$ is either increasing or decreasing At $x=2, f(2)=2^{3}+3(2)-2=12$ At $x=3, f(3)=3^{3}+3(3)-2=34$ $\therefore f(x) \in[12,34]$
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