Question
Show that the function $f(x)=x^3-3 x^2+6 x-100$ is increasing on $R.$
✓
Answer
Given, $f(x)=x^3-3 x^2+6 x-100$
Therefore, on differentiating both sides $w.r.t.\ x$, we get,
$ f^{\prime}(x)=3 x^2-6 x+6$
$=3 x^2-6 x+3+3$
$=3\left(x^2-2 x+1\right)+3$
$=3(x-1)^2+3>0$
$\therefore f^{\prime}(x)>0 $
This shows that function $f(x)$ is increasing on $R$.
Therefore, on differentiating both sides $w.r.t.\ x$, we get,
$ f^{\prime}(x)=3 x^2-6 x+6$
$=3 x^2-6 x+3+3$
$=3\left(x^2-2 x+1\right)+3$
$=3(x-1)^2+3>0$
$\therefore f^{\prime}(x)>0 $
This shows that function $f(x)$ is increasing on $R$.
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