Recent content by alex123aaa

could you tell me at least what is that formula please I am running out of time i have to turn this paper in tomorrow

here's my solution i really don't know if this is correct but please correct me so take the derivative = 4x^3-6x+2 Set it equal to zero and begin solving. 4x^3-6x+2 = 0 4x^3-6x = -2 2x(2x^2-3) = -2 2x^2 - 3 = 0 | 2x = 0 2x^2 = 3 | x = 0 x^2 = 3/2 x = plus or minus the square root of...

yeah i have to solve for x but how wil i get rid of the derivative 4x^3? i don't know how to remove it

how can i find the x-coordinates of all local extrema in this equations please anyone could answer it for me? f(x)=x^3+4x^2+2x f(x)=x^4-3x^2+2x f(x)=x^5-2x^2-4x f(x)=x^5+4x^2-4x

how can i find the local extrema of x-coordinates in this equation? f(x)=x^4-3x^2+2x