If f:[2,3] rightarrow R is defined by f(x)=x^ | vedclass

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(B) Given,$f(x)=x^3+3x-2$. On differentiating with respect to $x$,we get $f'(x)=3x^2+3$. Since $x^2 \ge 0$ for all $x \in [2,3]$,$f'(x) = 3x^2+3 \ge 3

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$[12,34]$

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(B) Given,$f(x)=x^3+3x-2$. On differentiating with respect to $x$,we get $f'(x)=3x^2+3$. Since $x^2 \ge 0$ for all $x \in [2,3]$,$f'(x) = 3x^2+3 \ge 3 > 0$. Thus,$f(x)$ is a strictly increasing function on the interval $[2,3]$. For an increasing function,the range is $[f(2), f(3)]$. At $x=2$,$f(2) = 2^3 + 3(2) - 2 = 8 + 6 - 2 = 12$. At $x=3$,$f(3) = 3^3 + 3(3) - 2 = 27 + 9 - 2 = 34$. Therefore,the range of $f(x)$ is $[12, 34]$.

If $f:[2,3] \rightarrow R$ is defined by $f(x)=x^3+3x-2$,then the range of $f(x)$ is contained in the interval

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