critical points

[Ryan Rigdon]

my f(x) = 4x^5 + 5x^3

f'(x) = 20x^4 + 15x^2

now my next part calls for critical points of f'(x). so i set f'(x) = 0 and solve for x.

20x^4 + 15x^2 = 0

5x^2(4x^2 + 3) = 0

5x^2 = 0 4x^2 + 3 = 0
x = 0 x^2 = -3/4
X = + or - Sqroot (-3/4) = + or - Sqroot (-3)/2

now would my cp's be x = 0, + or - (isqroot(3))/2

do my cp's seems to be ok or should i look at them again?

 

[tkhunny]

Sorry. x = 0 is the only Real solution. Take a good look at that other factor.