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

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barce90 | Student, Undergraduate

Posted April 14, 2013 at 12:40 AM via web

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

Iincreasing=

Decreasing=

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oldnick

Posted April 14, 2013 at 1:40 AM (Answer #1)

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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)`

 

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pramodpandey | College Teacher

Posted April 14, 2013 at 4:10 AM (Answer #2)

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`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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