Show that the slope of y = x^3 - 3x + 3 at any x is 3x^2 - 3 A=delta (dont know how to do delta) im doing it using this method ( (x+Ax)^3 - 3(x+Ax) + 3 ) - ( x^3 - 3x + 3 ) / ( (x+Ax) - x ) i simplify and get Ax^2 + 3xAx + 3x^2 - 6x
Follow Math Help Forum on Facebook and Google+
Originally Posted by stones44 Show that the slope of y = x^3 - 3x + 3 at any x is 3x^2 - 3 A=delta (dont know how to do delta) im doing it using this method ( (x+Ax)^3 - 3(x+Ax) + 3 ) - ( x^3 - 3x + 3 ) / ( (x+Ax) - x ) i simplify and get Ax^2 + 3xAx + 3x^2 - 6x where did 6x come from? you simplified something wrong. expanding everything, you should get: now cancel what is to be canceled, simplify as much as possible, and take the limit
Originally Posted by Jhevon where did 6x come from? you simplified something wrong. expanding everything, you should get: now cancel what is to be canceled, simplify as much as possible, and take the limit doesnt -3(x+Ax) = -3x + -3Ax ? and then the other -3x so -6x
let now, clean it up and finish finding the limit which will give you the derivative of f.
i figured what i did wrong ok so i cleaned it up i have 3x^2 + 3xAx + Ax^2 - 3 shouldnt i end up with (3x^2 - 3) +Ax ?
Originally Posted by stones44 doesnt -3(x+Ax) = -3x + -3Ax ? and then the other -3x so -6x what other -3x? the only other 3x is +3x
yeah i see what i did but i still dont have the answer do i? look at my post above yours
Originally Posted by stones44 yeah i see what i did but i still dont have the answer do i? look at my post above yours nope, you should end up with and if we let ...?
View Tag Cloud