Question

If \[f(x) = 3{x^4} + 4{x^3} - 12{x^2} + 12\] then \[f(x)\] is
A.Increasing in \[\left( { - \infty , - 2} \right)\] and in \[\left( {1,\infty } \right)\]
B.Increasing in \[\left( { - 2,0} \right)\] and in \[\left( {1,\infty } \right)\]
C.Decreasing in \[\left( { - 2,0} \right)\] and in \[\left( {0,1} \right)\]
D.Decreasing in \[\left( { - \infty , - 2} \right)\] and in \[\left( {1,\infty } \right)\]

Answer 1

Hint: Firstly find the derivative of the function as it can be at times used to determine whether a function is increasing or decreasing on any interval in its domain. If \[f'(x) > 0\] in an interval I, then the function is said to be increasing on I and if \[f'(x) < 0\] in an interval I, then the function is said to be decreasing on I.

Complete step-by-step answer:
If \[f'(x) > 0\] then \[f\] is increasing on the interval, and if \[f'(x) < 0\] then \[f\] is decreasing on the interval.
Following steps are involved in the process of finding the intervals of increasing and decreasing function:
Firstly, differentiate the given function with respect to the constant variable.
Then solve \[f'(x) = 0\] .
We are given \[f(x) = 3{x^4} + 4{x^3} - 12{x^2} + 12\]
Differentiating both the sides with respect to \[x\] we get ,
 \[f'(x) = 12{x^3} + 12{x^2} - 24x\]
Putting \[f'(x) = 0\] i.e. \[12{x^3} + 12{x^2} - 24x = 0\]
On simplification we get \[12x({x^2} + x - 2) = 0\]
Now splitting the middle term in the quadratic equation in bracket above we get
 \[12x({x^2} + 2x - x - 2) = 0\]
 \[12x\left( {x\left( {x + 2} \right) - 1\left( {x + 2} \right)} \right) = 0\]
Hence we get \[12x(x - 2)(x + 1) = 0\]
Hence \[x = 0, - 1,2\]
Hence we get the intervals \[\left( { - \infty , - 1} \right),\left( { - 1,0} \right),\left( {0,2} \right),\left( {2,\infty } \right)\]
In the interval \[\left( { - \infty , - 1} \right)\] \[f'(x) < 0\] hence decreasing.
In the interval \[\left( { - 1,0} \right)\] \[f'(x) > 0\] hence increasing.
In the interval \[\left( {0,2} \right)\] \[f'(x) < 0\] hence decreasing.
In the interval \[\left( {2,\infty } \right)\] \[f'(x) > 0\] hence increasing.
Therefore \[f(x)\] is increasing in \[\left( { - 1,0} \right)\] and \[\left( {2,\infty } \right)\] and \[f(x)\] is decreasing in \[\left( { - \infty , - 1} \right)\] and \[\left( {0,2} \right)\] .
Therefore \[f(x)\] is Increasing in \[\left( { - 2,0} \right)\] and in \[\left( {1,\infty } \right)\] .
So, the correct answer is “Option B”.

Note: After solving the equation of the first derivative and finding the points of discontinuity we get the open intervals with the value of \[x\] , through which the sign of the intervals can be taken into consideration.
If the sign of the interval in their first derivative form gives more than \[0\] then the function is said to be increasing in nature, while if the sign of the intervals in their first derivative form gives less than \[0\] then the function is said to be decreasing in nature.