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
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
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= $\displaystyle \pm 2$