f'=x^3-3x^2-10x=0
x(x^2-3x-10)=0
x(x-5)(x+2)=0
x=0;5;-2
Now you have to test each to see if it is a minimum
f"=3x^2-6x-10
at x=0, it is negative, so it is a max
at x=5, 75-30-10 is positive, a relative min
at x=-2
f"=12+12-10, a rel minimum
Find the relative minimum point(s) of
f(x)=x^4/4 - x^3 -5x^2 - 10
I know you have to take the derivative but I am not sure how to do that for this problem.
2 answers
oops, at x=-2, f" is negative, so a relative max.