Find the intervals in which the function f given by f(x)=4x^3-6x^2-72x+30 is (a) strictly increasing (b) strictly decreasing.

 

 

 

 
 
 
 
 

Answers (1)

f(x)=4x^3-6x^2-72x+30

Calculating f(x)

f'(x)=12x^2-12x-72

f'(x)=12\left (x^2-x-6 \right )

f'(x)=12\left (x^2-3x+2x-6 \right )

f'(x)=12\left (x(x-3)+2(x-3) \right )

f'(x)=12(x+2)(x-3)

Putting f'(x)=12(x+2)(x-3)=0

12(x+2)(x-3)=0

(x+2)(x-3)=0

x=-2\: \: and\: \: x=3

Plotting points on the number line

Value of x Intervals Sign of f'(x)=12(x+2)(x-3) Nature of function f
x<-2 (-\infty ,-2) (+)(-)(-)=(+)(+)=(+)>0 f is strictly increasing
-2<x<3 (-2 ,3) (+)(+)(-)=(+)(-)=(-)<0

f is strictly decreasing

x>3 (3 ,\infty) (+)(+)(+)=(+)(+)=(+)>0

f is strictly increasing

(a) f is strictly increasing in (-\infty ,-2) and (3 ,\infty)

(b) f is strictly decreasing in (-2 ,3)

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions