Use the first derivative to determine where the function f(x)=6x-3x^1/3 is increasing and decreasing.

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Decreasing=

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`f(x)=6x-3x^(1/3)` `f'(x)= 6-x^(-2/3)`

`f'(x)= 0` implies: `1/6= x^(2/3)`

powering both sides by `3/2` : `(1/6)^(3/2)=(x^(2/3))^(3/2)` `sqrt((1/6)^3)=x`

`(1/6)sqrt(1/6)= x` `x=sqrt(6)/36`

`f''(x)= -(2/3) (x)^(5/3)=-(2/3)x..^3sqrt(x^2)`

Is clear that : `f''(sqrt(6)/36)<0` then `x=sqrt(6)/36 ` isa max point

So that , function increases for `x<sqrt(6)/36` and decreases for `x>sqrt(6)/36` ,

while for `x= sqrt(6)/36` has the max value `sqrt(6)/6 - (sqrt(6)/6)^(1/3)`

`f(x)=6x-3x^(1/3)`

differentiate w.r.t x

`f'(x)=6-x^(-2/3)`

If `6x^(2/3)-1>0` then increasing otherwise decreasing.

`x^(2/3)>1/6`

`x^2>(1/6)^3`

`i.e. x>(1/6)^(3/2) or x<-(1/6)^(3/2)`

Thus function is increasing if `x in(-oo,-(1/6)^(3/2))uu((1/6)^(3/2),oo)`

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