# Gradient problems URGENT

• Jul 24th 2012, 07:19 AM
terone71
Find the coordinates of the points on the curve y=x^3 where the gradient is 12

P.S Can people explain this in detail as I am horrible at maths
• Jul 24th 2012, 07:41 AM
pratique21
Re: Gradient problems URGENT
Gradient means inclination and inclination points to something like 'Slope'
Now, what is slope ?
It is simply $\frac{dy}{dx}$ for any function y=f(x)
so here you have f(x)=x3
$\frac{dy}{dx} = 3x^2$
$x^2 = 4$
$x=+4, -4$
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 AM
terone71
Re: Gradient problems URGENT
Quote:

Originally Posted by pratique21
Gradient means inclination and inclination points to something like 'Slope'
Now, what is slope ?
It is simply $\frac{dy}{dx}$ for any function y=f(x)
so here you have f(x)=x3
$\frac{dy}{dx} = 3x^2$
$x^2 = 4$
$x=+4, -4$
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

Thanks. This some has been bothering me for some time!!!!
• Jul 24th 2012, 07:55 AM
terone71
Re: Gradient problems URGENT
Quote:

Originally Posted by pratique21
Gradient means inclination and inclination points to something like 'Slope'
Now, what is slope ?
It is simply $\frac{dy}{dx}$ for any function y=f(x)
so here you have f(x)=x3
$\frac{dy}{dx} = 3x^2$
$x^2 = 4$
$x=+4, -4$
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

One more question! How did you get x^2 = 4 in the first place? After getting dy/dx = 3x^2 . You then lost me!
• Jul 24th 2012, 08:05 AM
pratique21
Re: 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 PM
terone71
Re: 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 PM
pratique21
Re: Gradient problems URGENT
okay, now take the entire thing to one side (i.e subtract 4 from both sides)
so $x^2 = 4$
become: $x^2 - 4 =0$
now use $a^2 - b^2 = (a+b) \cross (a-b)$
so you have:- $(x-2) \cross (x+2)=0$
this is true when either $(x-2)= 0$ or $(x+2)=0$ or both are zero.
So x= $\pm 2$
• Jul 29th 2012, 03:31 AM
terone71
Re: Gradient problems URGENT
Thanks dood