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

pramodpandey | Student


differentiate w.r.t x


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



`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)`


oldnick | Student

`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)`