Answer:
f(x) = x³ +4x²+10
f'(x) = 3x²+8x
the critical point is that x value that has zero to be the first derivative function
3x²+8x = 0
when we factorize
x(3x+8) = 0
x = 0
3x+8 = 0
3x = -8
x = -8/3
at critical point when f"(x)>0, that is our minima
when f"<0 that is the maxima
f'(x) = 3x²+8x
f''(x) = 6x + 8
f"(0) = 8>0
the point x = 0 is minima point
f"(x) = -8/3 = -8<0 is Maxima
x coordinate of maxima = -8/3
