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Question: The function \(f\left( x \right)={{x}^{3}}-3x\) is, A. Increasing on \(\left( -\infty ,-1 \right)\...

The function f(x)=x3−3xf\left( x \right)={{x}^{3}}-3x is,
A. Increasing on (−∞,−1)∪(1,∞)\left( -\infty ,-1 \right)\cup \left( 1,\infty \right) and decreasing on (−1,1)\left( -1,1 \right).
B. Decreasing on (−∞,−1)∪(1,∞)\left( -\infty ,-1 \right)\cup \left( 1,\infty \right) and increasing on (−1,1)\left( -1,1 \right).
C. Increasing on (0,∞)\left( 0,\infty \right) and decreasing on (−∞,0)\left( -\infty ,0 \right).
D. Decreasing on (0,∞)\left( 0,\infty \right) and increasing on (−∞,0)\left( -\infty ,0 \right).

Explanation

Solution

Hint: A function y=f(x)y=f\left( x \right) is increasing when dydx>0\dfrac{dy}{dx}>0 and decreasing when dydx<0\dfrac{dy}{dx}<0. For f(x)=x3−3xf\left( x \right)={{x}^{3}}-3x, calculate f′(x)f'\left( x \right) and check where f′(x)>0f'\left( x \right)>0 and where f′(x)<0f'\left( x \right)<0. The interval where f′(x)>0,f(x)f'\left( x \right)>0,f\left( x \right) will be increasing and the interval where f′(x)<0,f(x)f'\left( x \right)<0,f\left( x \right) will be decreasing.

Complete step-by-step solution -
Given f(x)=x3−3xf\left( x \right)={{x}^{3}}-3x, we have to find the interval where f(x)f\left( x \right) is increasing and the interval where f(x)f\left( x \right) is decreasing.
We know that,
For a function f(x)f\left( x \right) to be increasing in an interval, its derivative f′(x)f'\left( x \right) should be positive in that interval and for a function f(x)f\left( x \right) to be decreasing in an interval its derivative f′(x)f'\left( x \right) should be negative in the interval.
Let us first find the derivative of f(x)f\left( x \right),
f(x)=x3−3x f′(x)=3x2−3 [As we know that derivative of xn=nxn−1] \begin{aligned} & f\left( x \right)={{x}^{3}}-3x \\\ & f'\left( x \right)=3{{x}^{2}}-3 \\\ & \left[ \text{As}\ \text{we}\ \text{know}\ \text{that}\ \text{derivative}\ \text{of}\ {{x}^{n}}=n{{x}^{n-1}} \right] \\\ \end{aligned}
Let us find the region where f′(x)>0f'\left( x \right)>0;
f′(x)>0 ⇒(3x2−3)>0 \begin{aligned} & f'\left( x \right)>0 \\\ & \Rightarrow \left( 3{{x}^{2}}-3 \right)>0 \\\ \end{aligned}
Dividing both sides by ‘3’, we will get,
⇒(x2−1)>0\Rightarrow \left( {{x}^{2}}-1 \right)>0
Factorising (x2−1)=(x−1)(x+1)\left( {{x}^{2}}-1 \right)=\left( x-1 \right)\left( x+1 \right) using ′′a2−b2=(a−b)(a+b)′′''{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)'', we will get,
⇒(x−1)(x+1)>0\Rightarrow \left( x-1 \right)\left( x+1 \right)>0
Sign diagram for this equation,

So, at x=−1 and x=1 f′(x)=0x=-1\ and\ x=1\ f'\left( x \right)=0
And for x∈(−∞,−1)∪(1,∞),f′(x)>0x\in \left( -\infty ,-1 \right)\cup \left( 1,\infty \right),f'\left( x \right)>0
And for x∈(−1,1),f′(x)<0x\in \left( -1,1 \right),f'\left( x \right)<0
Thus, f(x)f\left( x \right) is increasing in interval (−∞,−1)∪(1,∞)\left( -\infty ,-1 \right)\cup \left( 1,\infty \right) and decreasing in the interval (−1,1)\left( -1,1 \right).
And option (A) is the correct answer.

Note: Be careful that x=−1 and x=1 x=-1\ and\ x=1\ won’t be included in either increasing interval or decreasing interval. There will be either a local maxima or local minima at x=1 and x=−1  as f′(x)=0x=1\ and\ x=-1\ \ as\ f'\left( x \right)=0 at x=1 and x=−1 x=1\ and\ x=-1\ .