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#f(x)=x^3-3x-x^2-x+10#
I am not sure why you posted the problem in this form because it simplifies to:
#f(x)=x^3-x^2-4x+10#. To find the minimum and maximum of a function we take the derivative of it, set it equal to zero and solve for its roots. Then we plug those roots in the original function. One of them will produce a higher #y# value and one lower.
The higher one is the maximum and the lower one the minimum:
#f'(x)=3x^2-2x-4#
#3x^2-2x-4=0#
#x=(2+-sqrt(4-4(3)(-4)))/6=(2+-sqrt52)/6=(2+-2sqrt13)/6#
#x=(1+-sqrt13)/3#
#f((1+sqrt13)/3)=((1+sqrt13)/3)^3-((1+sqrt13)/3)^2-4((1+sqrt13)/3)+10=5.1206# minimum
#f((1-sqrt13)/3)=12.0646# maximum
See the graph:
graph{x^3-x^2-4x+10 [-80, 80, -40, 40]}
