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find the intervals of which the the function  f(x)= 3x4-4x3-12x2+5 is  (a) strictly increasing                      (b) strictly decrasing

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fx=3x4-4x3-12x2+5Differentiate both sides with respect to x, we get:f'x=12x3-12x2-24x=12xx2-x-2=12xx2-2x+x-2=12xxx-2+1x-2⇒f'x=12xx-2x+1By method interval we get f'x>0 when x∈-1,0∪2,∞, hence fx is strictly increasing in x∈-1,0∪2,∞f'x<0 when x∈-∞,-1∪0,2, hence fx is strictly decreasing in x∈-∞,-1∪0,2
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f (x)=3x​4 - 4x3 - 12x2 + 5
differentiating w.r.t x,
f ' (x)= 12x3 - 12x2 - 24x

let f ' (x)=0
12x3 - 12x2 - 24x = 0
x2 - x - 2 =0
x2 + x - 2x -2= 0
(x + 1) (x - 2) = 0

x= - 1  x = 2
the intervals are (- ∞,- 1), (- 1,2), (2,∞)

in (- ∞,- 1) take x= - 2
f ' (- 2)=144 + 48 -24
f ' (-2)>0
f  (x) is  strictly increasing in (- ∞​,- 1)

in interval (2,∞)
​take x = 3
f ' (3) = 324 -72 -24
f ' (3) > 0
f  (x) is strictly increasing in (2 , ∞)

in interval (- 1,2)
take x = 0
f ' (0) = -24
f ' (0) < 0
f  (x) is strictly decreasing in (- 1,2)

f (x) is strictly increasing in (-∞​ ,-1) U (2, ∞​)
f (x) is strictly decreasing in (- 1,2)
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