Patrick B. answered 08/05/20
Math and computer tutor/teacher
f'(5) < 0
the critical points are:
f'(-2) = 0
f'(2) = 0
So then
y' = (x+2)(2-x)
= 2x + 4 - x^2 - 2x
= 4 - x^2
integrating:
y = 4x - (1/3)x^3 + c
you can graph this function and
it's derivative at
www dot desmos dot com
add a slider for c;
even for c=0, the function possesses
these properties
John C.
Can you explain the integrating part? I do not understand where (1/3)x^3 comes from10/07/20