What is the relative maximum and minimum of the function?
f(x)=x^3+6x^2-36
the relative maximum is at (–6, 216) and the relative minimum is at (2, –40).
the relative maximum is at (–6, 40) and the relative minimum is at (2, –216).
the relative maximum is at (6, 216) and the relative minimum is at (–2, –40).
the relative maximum is at (6, 40) and the relative minimum is at (–2, –216).

Take the derivitive
f'(x)=3x^2+12x
find where it equals zero
it equal zero at x=0 and x=-4
find the y values
f(0)=-36
f(-4)=-4
the critical points are (0,-36) and (-4,-4)

make sign chart
evalutat f'(x) at x=-5 and x=-1 and x=1, and see their signs
f'(-5)=(+)
f'(-1)=(-)
f'(1)=(+)
see below attachment

max is where sign changes from (+) to (-)
min is where sign changes from (-) to (+)

so
max at (-4,-4)
min at (0,-36)
none of the options are correct, do you have te right problem?

Zero is the mid point of the line segment