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)

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