Separate the intervals of monotonocity for the function `f(x)=-2x^3-9x^2-12 x+1`

Answer

(a) Increasing (-2,-1)
Decreasing : `(-oo,-2)cup(-1,oo)`
(b) Increasing : (0,2)
Decreasing : `(-oo,0)cup(2,oo)`
(c ) Increasing : `(0,pi//4)cup(5pi//4,2pi)`
Decreasing:`(pi//4,5pi//4)`
(d) Increasing :`(pi//4,5pi//4)`
Decreasing :`(0,pi//2)cup(2pi//3,pi)`
(e ) Increasing : `(1//sqrt(e,oo))`
Decreasing `(0,1//sqrt(e))`

Explanation

(a) `f(X)=22x^(3)-9x^(2)-12x+1`
`f(x)=-6x^(2)-18x-12`
`=-6(x+2)(x+1)`
or `f'(x)gt0 if x in (-2,-1)`
and `f'(x)lt0, if x in (-oo.-2)cup(-1,oo)`
Thus f(x) is increasing for x `in` (-2,1) and
f(X) is decreasing for x `in (-oo,2)cup(-1,oo)`
(b) Let y=f(x)=`x^(2)e^(-x)`
`therefore (dy)/(dx)=2xe^(-x)-x^(2)e^(-x)`
`=e^(-x)(2x-x^(2))`
`=e^(-x)x(2-x)`
f(x) is increasing if `f(x)gt0 or x (2-x)gt0 or x in (0,2)`
f(x) is decreasing if `f'(x)lt0 or x(2-x)lt0`
or `x in (-oo,0)cup(2,oo)`
(c ) we have f'(x) =cosx -sin x

f(x) is increasing if `f'(x) gt0 or cos x gt sin x `
or `x in (0,pi//4)cup ((5pi)/(4),2pi)` , (see the graph)
f(x) is decreasing if `f'(x) lt 0 or cos x lt sinx `
or`x in (pi//4,(5pi)/(4))`
(d) Given f(x) =`3cos^(4)x+10 cos^(3)x+6cos^(2)x-3`
`therefore f'(x) =12 cos^(3)x(-sin x)+30 cos^(2)x(-sin x) + 12 cosx-(-sinx)`
`=-3 sin 2x(2 cos^(2) x+5 cos x +2)`
`=-3 sin 2x(2 cos x+1)(cos x+2)`
`f'(x) =0 rarrr sin 2x =0 rarr 0,(pi)/(2), pi`
or `2cos x +1 =0 rarr x=(2pi)/(3)`
as `cos x+ 2 ne 0`
sign scheme of f'(x) is as follow
So , f (x) decrease on `(0,(pi)/(2))cup((2pi)/(3),pi)` and incerase on `((pi)/(2),(2pi)/(3))`
(e ) f(X) =`(log_(e)x)^(2)+(log_(e)x),xgt0`
`therefore f'(x) =2(log_(e)x)/(x)+(1)/(x)=(2log_(e)x+1)/(x)`
f(X) increasing when 2 `log_(e)x+1gt0`
or `log_(e)xgt-(1)/(2)`
or `xgte^(-1)/(2)`
or f(x) increases when`x in ((1)/sqrt(e),oo)`
or f(x) decreases when x `in (0,(1)/sqrt(e))`