Find the coordinates of the points on the curve y=x^3 where the gradient is 12

Please help URGENT

P.S Can people explain this in detail as I am horrible at maths

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- Jul 24th 2012, 07:19 AMterone71Gradient problems URGENT
Find the coordinates of the points on the curve y=x^3 where the gradient is 12

Please help URGENT

P.S Can people explain this in detail as I am horrible at maths - Jul 24th 2012, 07:41 AMpratique21Re: Gradient problems URGENT
Gradient means inclination and inclination points to something like 'Slope'

Now, what is slope ?

It is simply $\displaystyle \frac{dy}{dx} $ for any function y=f(x)

so here you have f(x)=x^{3}

$\displaystyle \frac{dy}{dx} = 3x^2 $

$\displaystyle x^2 = 4 $

$\displaystyle x=+4, -4 $

Now you have the value of x. To find the complete set (x,y) just plug the values of x in to f(x)

f(4)=64

f(-4)=-64 - Jul 24th 2012, 07:52 AMterone71Re: Gradient problems URGENT
- Jul 24th 2012, 07:55 AMterone71Re: Gradient problems URGENT
- Jul 24th 2012, 08:05 AMpratique21Re: Gradient problems URGENT
i put the value of dy/dx that is given as 12

so cancelling 3 gives me x^2 = 4 - Jul 24th 2012, 04:19 PMterone71Re: Gradient problems URGENT
But then after that how did you still find x? If you cancel 3 then its x^2=4

- Jul 24th 2012, 09:14 PMpratique21Re: Gradient problems URGENT
okay, now take the entire thing to one side (i.e subtract 4 from both sides)

so $\displaystyle x^2 = 4$

become: $\displaystyle x^2 - 4 =0$

now use $\displaystyle a^2 - b^2 = (a+b) \cross (a-b)$

so you have:- $\displaystyle (x-2) \cross (x+2)=0 $

this is true when either $\displaystyle (x-2)= 0 $ or$\displaystyle (x+2)=0 $ or both are zero.

So x= $\displaystyle \pm 2$ - Jul 29th 2012, 03:31 AMterone71Re: Gradient problems URGENT
Thanks dood