​Show that function f(x)
x ³ – 3x ² + 6x – 100 increasing of R

Dear student,
$$\begin{gathered} f\left(x\right) = x^3-3x^2+6x-100 \\ differentiate \\ f\text{'}\left(x\right) = 3x^2-6x+6 \\ for increa\sin g funtion in R prove f\text{'}\left(x\right)>0 \\ \mathit{p}\mathit{r}\mathit{o}\mathit{o}\mathit{f} \\ f\text{'}\left(x\right) = 3x^2-6x+6>0 \\ D=b^2-4ac<0 for positve function in x\in R \\ D=36 - 4\times 6\times 3 \\ D = -36 \\ D<0 \\ so D<0 then f\text{'}\left(x\right) >0 in x\in R \\ then f\left(x\right) is increa\sin g funtion in R \end{gathered}$$
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