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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`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)`

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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