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Prove that the function given by $f(x)=x^3-3 x^2+3 x-100$ is increasing in $R$.
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Verified Answer
$f^{\prime}(x)=3 x^2-6 x+3=3\left(x^2-2 x+1\right)=3(x-1)^2$
Now $\mathrm{x} \in \mathrm{R}, \quad \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}-1)^2 \geq 0$
i.e. $\mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \forall \mathrm{x} \in \mathrm{R}$; hence, $\mathrm{f}(\mathrm{x})$ is increasing on $\mathbf{R}$.
Now $\mathrm{x} \in \mathrm{R}, \quad \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}-1)^2 \geq 0$
i.e. $\mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \forall \mathrm{x} \in \mathrm{R}$; hence, $\mathrm{f}(\mathrm{x})$ is increasing on $\mathbf{R}$.
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