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

2. ## Re: 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)=x3
$\displaystyle \frac{dy}{dx} = 3x^2$
$\displaystyle x^2 = 4$
$\displaystyle x=+4, -4$
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

3. ## Re: Gradient problems URGENT

Originally Posted by pratique21
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)=x3
$\displaystyle \frac{dy}{dx} = 3x^2$
$\displaystyle x^2 = 4$
$\displaystyle x=+4, -4$
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
Thanks. This some has been bothering me for some time!!!!

4. ## Re: Gradient problems URGENT

Originally Posted by pratique21
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)=x3
$\displaystyle \frac{dy}{dx} = 3x^2$
$\displaystyle x^2 = 4$
$\displaystyle x=+4, -4$
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
One more question! How did you get x^2 = 4 in the first place? After getting dy/dx = 3x^2 . You then lost me!

5. ## Re: Gradient problems URGENT

i put the value of dy/dx that is given as 12
so cancelling 3 gives me x^2 = 4

6. ## Re: Gradient problems URGENT

But then after that how did you still find x? If you cancel 3 then its x^2=4

7. ## Re: 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= $\displaystyle \pm 2$

Thanks dood