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Use the first derivative to determine where the function f(x)=6x^4+192x is increasing...

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elpony | Student, Undergraduate | eNoter

Posted April 14, 2013 at 12:29 AM via web

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Use the first derivative to determine where the function

f(x)=6x^4+192x is increasing and decreasing.

hint: X^3+a^3=(x+a)(x^2-ax+a^2)

increasing:

decreasing:

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oldnick | Valedictorian

Posted April 14, 2013 at 1:11 AM (Answer #1)

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`f(x)=6x^4+192x=6x(x^3-32)`   (1)

`f'(x)= 24x^3+192`

`f'(x)=24x^3+192=0`     dividing by 24:

`x^3+8=0`      then    `x=-2` 

`f''(x)=3x^2`    and `f''(-2)=12 >0`  

Thus `x= -2`  is a min. point, it means function decrease  for `x< -2`    and decrease for `x>2`   assuming the value  `y=-288`  for  `x=-2`

 

by the  (1) we know has zero for `x= 0`   and  `x=-2 ^3sqrt(4)`

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pramodpandey | College Teacher | Valedictorian

Posted April 14, 2013 at 4:22 AM (Answer #2)

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f(x)=6x^4+192x

`f(x)=6x^4+192x`

`f'(x)=24x^3+192`

`f'(x)=8(x^3+8)`

`f'(x)=8(x^3+2^3)`

Function is an increasing function if

`x^3+2^3>0`

`(x+2)(x^2+2^2-2x)>0```

`x+2>0 and x^2+4-2x>0`

since f is real function so ,only possible `x>2`

Thus function is increasing if `x in(2,oo)` and decreasing is `x in(-oo,2)`

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