f(x) = 3x^4-8x^3-90x^2+50f(x)=3x4−8x3−90x2+50
"dom"(f) = (-oo,oo)dom(f)=(−∞,∞)
f'(x) = 12x^3-24x^2-180x
f'(x) is never undefined
f'(x) = 12x(x^2-2x-15) = 12x(x-5)(x+3) = 0 at x= -3, 0, 5
Here is one version of the sign chart for f'(x)
{: (bb"Intervals:",(-oo,-3),(-3,0),(0,5),(5,oo)),
(darr bb"Factors"darr,"========","======","=====","======"),
(12x, bb" -",bb" -",bb" +",bb" +"),
(x-5,bb" -",bb" -",bb" -",bb" +"),
(x+3,bb" -",bb" +",bb" +",bb" +"),
("==========","========","======","=====","======"),
(bb"Product"=f'(x),bb" -",bb" +",bb" -",bb" +")
:}
At x=-3, the sign of f'(x) changes from - to +, (so f changes from decreasing to increasing) so f(-3) = -301 is a local minimum.
At x=0, the sign of f'(x) changes from + to -, (so f changes from increasing to decreasing) so f(0) = 50 is a local maximum.
At x=5, the sign of f'(x) changes from - to +, (so f changes from decreasing to increasing) so f(5) = -1325 is a local minimum.
