Question

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

Answer 1

Hint: 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 $ .