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