#f(x) = x^3+5x^2#
Apply the power rule.
#f'(x) =3x^2+10x#
#f(x)# will be increasing where #f'(x)>0#
and decreasing where #f'(x)<0#
Hence, #f(x)# is increasing where:
#3x^2+10x>0#
#x(3x+10)>0 ->#
#x<0 and (3x+10)<0 or x>0 and (3x+10)>0#
Note: If #x>0# then #(3x+10)# must be #>0#
#x<0 and (3x+10)<0 -> -oo < x < -10/3#
and
#x>0 -> 0< x < +oo#
Hence #f(x)# is increasing for #x in (-oo,-10/3)uu(0,+oo)#
#f(x)# is decreasing where:
#3x^2+10x<0#
#x(3x+10)<0 ->#
#x<0 and (3x+10)>0 or x>0 and (3x+10)<0#
Note: If #x>0# then #(3x+10)# cannot be #<0#
#:. x<0 and (3x+10)>0 -> -10/3 < x < 0#
Hence, #f(x)# is decreasing for #x in (-10/3, 0)#
#fx)# will have turning points where #f'(x)=0#
I.e. where: #x(3x+10)=0#
#:.x = -10/3 or 0#
Since #f(x)# is decreasing for #x in (-10/3,0)# and increasing thereafter, it is clear that:
#f(-10/3) = f_max and f(0) = f_min#
(This can also be verified by the 2nd derivative test if required)
Hence, #f(x)# has a local maximum at #x=-10/3# and a local minimum at #x=0#
These results can be observed from the graph of #f(x)# below.
graph{x^3+5x^2 [-41.1, 41.1, -20.53, 20.58]}
