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, 08:19 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, 08:41 AMpratique21Re: Gradient problems URGENT
Gradient means inclination and inclination points to something like 'Slope'

Now, what is slope ?

It is simply for any function y=f(x)

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

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, 08:52 AMterone71Re: Gradient problems URGENT
- Jul 24th 2012, 08:55 AMterone71Re: Gradient problems URGENT
- Jul 24th 2012, 09:05 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, 05:19 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, 10:14 PMpratique21Re: Gradient problems URGENT
okay, now take the entire thing to one side (i.e subtract 4 from both sides)

so

become:

now use

so you have:-

this is true when either or or both are zero.

So x= - Jul 29th 2012, 04:31 AMterone71Re: Gradient problems URGENT
Thanks dood