L lauren52 New member Joined Oct 28, 2008 Messages 6 Oct 28, 2008 #1 I am having trouble factoring and finding the critical points. g(x) = 3x^4+8x^2-24 g'(x) = 12x^3+16x = 12x^3+16x=0 = 4x(3x^2+4)=0 4x+0 I am unsure of the next step regarding what's left in parenthesis (3x^2+4)
I am having trouble factoring and finding the critical points. g(x) = 3x^4+8x^2-24 g'(x) = 12x^3+16x = 12x^3+16x=0 = 4x(3x^2+4)=0 4x+0 I am unsure of the next step regarding what's left in parenthesis (3x^2+4)
D Deleted member 4993 Guest Oct 28, 2008 #2 Re: Generalized power rule lauren52 said: I am having trouble factoring and finding the critical points. g(x) 3x^4+8x^2-24 g'(x) 12x^3+16x 12x^3+16x=0 4x(3x^2+4)=0 4x+0 I am unsure of the next step regarding what's left in parenthesis (3x^2+4) Click to expand... Your one solution is 4x = 0 x = 0 other one 3x^2 + 4 = 0 x^2 = - 4/3 <<<<<< no solution becuse a square is alsways positive in real domain.
Re: Generalized power rule lauren52 said: I am having trouble factoring and finding the critical points. g(x) 3x^4+8x^2-24 g'(x) 12x^3+16x 12x^3+16x=0 4x(3x^2+4)=0 4x+0 I am unsure of the next step regarding what's left in parenthesis (3x^2+4) Click to expand... Your one solution is 4x = 0 x = 0 other one 3x^2 + 4 = 0 x^2 = - 4/3 <<<<<< no solution becuse a square is alsways positive in real domain.
L lauren52 New member Joined Oct 28, 2008 Messages 6 Oct 28, 2008 #3 Re: Generalized power rule Thank you
