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)`
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.
To simplify, divide both sides by the GCF which is 56.
Note that we may re-write 27 as `3^3` . So, we have a sum of two cubes.
Then, factor left side. Apply the hint given above.
Then, set each factor equal to zero and solve for x.
For the first factor,
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` .