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If $f:[2,3] \rightarrow R$ 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:
$[12,34]$
Given, $f(x)=x^3+3 x-2$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
& \quad f^{\prime}(x)=3 x^2+3 \\
& \text { Put } \quad f^{\prime}(x)=0 \Rightarrow 3 x^2+3=0 \\
& \Rightarrow \quad x^2=-1 \\
& \therefore \quad f(x) \text { is either increasing or decreasing. } \\
& \text { At } x=2, f(2)=2^3+3(2)-2=12 \\
& \text { At } x=3, f(3)=3^3+3(3)-2=34 \\
& \therefore f(x) \in[12,34] .
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
& \quad f^{\prime}(x)=3 x^2+3 \\
& \text { Put } \quad f^{\prime}(x)=0 \Rightarrow 3 x^2+3=0 \\
& \Rightarrow \quad x^2=-1 \\
& \therefore \quad f(x) \text { is either increasing or decreasing. } \\
& \text { At } x=2, f(2)=2^3+3(2)-2=12 \\
& \text { At } x=3, f(3)=3^3+3(3)-2=34 \\
& \therefore f(x) \in[12,34] .
\end{aligned}$
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