Question

Prove that the function $f(x)={{x}^{3}}-6{{x}^{2}}+12x-18$ is strictly increasing on $\mathbb{R}$ .

Answer 1

Analyze the first order derivative $f'(x)$ of the given function.
If $f'(x)>0$ in a given region, then the function is increasing in that region.
Recall that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ .
Remember that the even powers of a real number are never negative. In particular, ${{n}^{2}}\ge 0,\forall n\in \mathbb{R}$ .

Complete step-by-step answer:
Let us differentiate the given function:
$f(x)={{x}^{3}}-6{{x}^{2}}+12x-18$
Using $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ , we get:
⇒ $f'(x)=3{{x}^{2}}-12x+12$
Separating the common factor 3 and factoring, we get:
⇒ $f'(x)=3\left( {{x}^{2}}-4x+4 \right)$
⇒ $f'(x)=3{{(x-2)}^{2}}$
Since ${{(x-2)}^{2}}\ge 0$ for any real value of x, $f'(x)=0$ at only one point $x=2$ , and $f'(x)>0$ for all $x\ne 0$ .
For a function which changes its direction from increasing to decreasing (or from decreasing to increasing), $f'(x)>0$ on one side and $f'(x)<0$ on the other side.
In this case $f'(x)>0,\forall x\in \mathbb{R}$ , therefore, the given function $f(x)$ is strictly increasing in $\mathbb{R}$ .

Note: A function $y=f(x)$ is increasing or decreasing if there is an increase or decrease in the value of $y$ when the value of $x$ is increased.
For a function $y=f(x)$ :
In the regions where f(x) is increasing, ${f}'(x)>0$ .
In the regions where f(x) is decreasing, ${f}'(x)<0$ .
At the points of relative (local) maxima or minima, ${f}'(x)=0$ .
At the points of relative (local) maxima, ${f}''(x)<0$ .
At the points of relative (local) minima, ${f}''(x)>0$ .