Question

Find the interval in which the function \[f(x)=3{{x}^{4}}-4{{x}^{3}}-12{{x}^{2}}+5\] is
A. Strictly increasing
B. Strictly decreasing

Answer 1

Hint: To solve this question, we will use the basic concept of a function to be increasing or decreasing which is given below.
A function g(x) is strictly increasing in an interval of its derivative g'(x) is strictly greater than 0 that is $g'(x)> 0$ then g(x) is strictly increasing in that interval. A function h(x) is strictly decreasing in an interval if it's derivative h'(x) is strictly less than 0.
We will put f'(x) = 0 to get required interval points and then check in all intervals if $f'(X) >0\Rightarrow f'(x) <0$

Complete step-by-step solution:
Given, \[f(x)=3{{x}^{4}}-4{{x}^{3}}-12{{x}^{2}}+5\]
A function g(x) is strictly increasing in an interval of its derivative g'(x) is strictly greater than 0 that is $g'(x)>0$ then g(x) is strictly increasing in that interval. A function h(x) is strictly decreasing in an interval if it's derivative h'(x) is strictly less than 0.
That is if $h'(x)<0$ then h(x) is strictly decreasing in that given interval.
We have, \[f(x)=3{{x}^{4}}-4{{x}^{3}}-12{{x}^{2}}+5\]
We have \[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\]
Using this above and differentiating f(x) with respect to x, we get
\[\begin{align}
  & \dfrac{d}{dx}f(x)=f'(x)=3\times 4{{x}^{3}}-4\times 3{{x}^{2}}-12\times 2x+0 \\
 & f'(x)=12{{x}^{3}}-12{{x}^{2}}-24x\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
To find interval of increasing or decreasing we will put f'(x) = 0 and try to get value of x possible when $f'(x) = 0$.
Putting $f'(x) = 0$
\[12{{x}^{3}}-12{{x}^{2}}-24x\text{ }=\text{ }0\]
Taking 12x common from above equation, we get
\[\begin{align}
  & 12\left( x \right)\left( {{x}^{2}}-x-2 \right)=0 \\
 & \Rightarrow 12x\left( {{x}^{2}}-x-2 \right)=0 \\
\end{align}\]
Splitting by middle term the equation $\left( {{x}^{2}}-x-2 \right)$ by using $-2x+x=-x$ we get
\[\begin{align}
  & 12x\left( {{x}^{2}}-2x+x-2 \right)=0 \\
 & 12x\left( x\left( x-2 \right)+1\left( x-2 \right) \right)=0 \\
 & \left( 12x \right)\left( x-2 \right)\left( x+1 \right)=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
\end{align}\]
So, we have three possibility of x.
\[\begin{align}
  & 12x=0\Rightarrow x=0 \\
 & x-2=0\Rightarrow x=2 \\
 & x+1=0\Rightarrow x=-1 \\
\end{align}\]
Hence, we will divide intervals using values of x obtained above.
The intervals are \[\left( -\infty ,-1 \right),\left( -1,0 \right),\left( 0,2 \right),\left( 2,\infty \right)\]
Make this interval on the real line as below:

Consider equation (i) \[f'(x)=12{{x}^{3}}-12{{x}^{2}}-24x\]
Also, \[f'(x)=\left( 12x \right)\left( x-2 \right)\left( x+1 \right)\] by equation (ii)
Case I - $x\in \left( -\infty ,-1 \right)$
When $x\in \left( -\infty ,-1 \right)$
Then \[\begin{align}
  & 12x=\text{negative} \\
 & x+1\text{=negative} \\
 & x-2\text{=negative} \\
\end{align}\]
So, product of all \[(12x)(x+1)(x-2)=(-)(-)(-)=(-)=\text{negative}\]
So, $f'(x)<0$ when $x\in \left( -\infty ,-1 \right)$
Case II - $x\in \left( -1,0 \right)$
When $x\in \left( -1,0 \right)$
Then \[\begin{align}
  & 12x=\text{negative} \\
 & x+1\text{=positive} \\
 & x-2\text{=negative} \\
\end{align}\]
So, product of all \[(12x)(x+1)(x-2)=(-)(+)(-)=(+)=\text{Positive}\]
So, $f'(x)>0$ when $x\in \left( -1,0 \right)$
Case III - $x\in \left( 0,2 \right)$
When $x\in \left( 0,2 \right)$
Then \[\begin{align}
  & 12x=\text{positive} \\
 & x+1\text{=positive} \\
 & x-2\text{=negative} \\
\end{align}\]
So product of all \[(12x)(x+1)(x-2)=(+)(+)(-)=(-)=\text{Negative}\]
So, $f'(x)<0$ when $x\in \left( 0,2 \right)$
Case IV - $x\in \left( 2,\infty \right)$
When $x\in \left( 2,\infty \right)$
Then \[\begin{align}
  & 12x=\text{positive} \\
 & x+1\text{=positive} \\
 & x-2\text{=positive} \\
\end{align}\]
So product of all \[(12x)(x+1)(x-2)=(+)(+)(+)=(+)=\text{positive}\]
So, $f'(x)>0$ when $x\in \left( 2,\infty \right)$
Hence, from Case I, II, III, IV we see that
$f'(x)>0$ when $x\in \left( -1,0 \right)\cup \left( 2,\infty \right)$ and $f'(x)<0$ when $x\in \left( -\infty ,-1 \right)\cup \left( 0,2 \right)$
So, f(x) is A: strictly increasing when $x\in \left( -1,0 \right)\cup \left( 2,\infty \right)$
And B: strictly decreasing when $x\in \left( -\infty ,-1 \right)\cup \left( 0,2 \right)$

Note: Always remember that, because we are considering $f'(x)>0\Rightarrow f'(x)<0$ and not considering $f'(x)\ge 0\Rightarrow f'(x)\le 0$ therefore, the end points like here end point are (0, 2, -1) cannot be included in the interval. That is, we will always have open interval (a, b) and not closed interval [a, b] used in such cases as we do not have $f'(x)\ge 0\Rightarrow f'(x)\le 0$