# Use the derivative to determine where the function `f(x)=14x^4+1512x` is increasing and decreasing Hint:`x^3+a^3=(x+a) (x^2-ax+a^2)`

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`f(x)=14x^4+1512x`

To determine the interval in which the function is increasing or decreasing, we need to determine the critical points. To do so, take the derivative of f(x).

`f'(x) = 56x^3 + 1512`

Then, set f'(x) equal to zero.

`0=56x^3+1512`

To simplify, divide both sides by the GCF which is 56.

`0/56=(56x^3+1512)/56`

`0=x^3+27`

Note that we may re-write 27 as `3^3` . So, we have a sum of two cubes.

`0=x^3+3^3`

Then, factor left side. Apply the hint given above.

`0=(x+3)(x^2-3x+9)`

Then, set each factor equal to zero and solve for x.

For the first factor,

`x+3=0`

`x=-3`

And for the second factor,

`x^2-3x+9 = 0`

use the quadratic formula to solve for x.

`x= (-b+- sqrt(b^2-4ac))/(2a)= (-(-3)+-sqrt((-3)^2-4(1)(9)))/(2*1) = (3+-sqrt(-27))/2 = (3+-3sqrt3i)/2`

Since the values of x in the second factor are imaginary, then,we consider only x=-3.

Substitute this value of x to f(x).

`f(x) = 14x^4+1512x=14(-3)^4+1512(-3)=-3402`

Hence, the critical point of f(x) is (-3,-3402).

Next, we need to assign a value of x that is less than the critical value -3 and substitute it to f'(x).

`x=-4` , `f'(x) = 56x^3+1512 = 56(-4)^3 +1512 = -2072`

*Since the sign of f'(x) is negative, this indicates that at x<-3, the function is decreasing*.

Also, assign a value of x that is greater than the critical values -3. Substitute this too to f'(x).

`x=0` , `f'(x)=56x^3+1512=56(0)^3+1512=1512`

*Since f'(x) is positive, this means that the function is increasing at x>-3*.

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**Therefore, the function `f(x) = 4x^4+1512x` is decreasing at `xlt-3` and increasing at `xgt-3` .**