Question
Question: If \[f(x) = 3{x^4} + 4{x^3} - 12{x^2} + 12\] then \[f(x)\] is A.Increasing in \[\left( { - \infty...
If f(x)=3x4+4x3−12x2+12 then f(x) is
A.Increasing in (−∞,−2) and in (1,∞)
B.Increasing in (−2,0) and in (1,∞)
C.Decreasing in (−2,0) and in (0,1)
D.Decreasing in (−∞,−2) and in (1,∞)
Solution
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)=3x4+4x3−12x2+12
Differentiating both the sides with respect to x we get ,
f′(x)=12x3+12x2−24x
Putting f′(x)=0 i.e. 12x3+12x2−24x=0
On simplification we get 12x(x2+x−2)=0
Now splitting the middle term in the quadratic equation in bracket above we get
12x(x2+2x−x−2)=0
12x(x(x+2)−1(x+2))=0
Hence we get 12x(x−2)(x+1)=0
Hence x=0,−1,2
Hence we get the intervals (−∞,−1),(−1,0),(0,2),(2,∞)
In the interval (−∞,−1) f′(x)<0 hence decreasing.
In the interval (−1,0) f′(x)>0 hence increasing.
In the interval (0,2) f′(x)<0 hence decreasing.
In the interval (2,∞) f′(x)>0 hence increasing.
Therefore f(x) is increasing in (−1,0) and (2,∞) and f(x) is decreasing in (−∞,−1) and (0,2) .
Therefore f(x) is Increasing in (−2,0) and in (1,∞) .
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.