Find the x-coordinates of all local maxima using the second derivative test. f(x)=x3+4x2+10 If there are multiple values, give them separated by commas. If there are no local maxima, enter ∅.

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

b.an increase in a causes b to decrease