Finding the two points on the curve $y=x^3 + 3x + 4$ at which the gradient is $6$ I already was able to find out the first coordinate, which is $(1,8)$; however I can't seem to find the second one.
 A: Any sort of question that involves finding the gradient at a certain point should give you the signal to use differentiation.
Differentiating our function with respect to $x$ then gives $$\frac{dy}{dx}=3x^2+3.$$ Now, they've given you the gradient, namely $6$, so we set our derivative equal to this and we get $$3x^2+3=6\Rightarrow x=\pm 1.$$ Finally, we must substitute our values of $x$ into our original function to obtain the $y$ values. So, we get $$y=1^3+3(1)+4=8,\quad y=(-1)^3+3(-1)+4=0.$$ So the points are $(1,8)$ and $(-1,0)$.
A: The derivative of $x^3+3x+4$ with respect to $x$ is $3x^2+3$. This derivative gives the instantaneous slope of the original function as a function of $x$. 
Since you are looking for the point at which the slope of $x^3+3x+4$ is $6$, just set $3x^2+3$ equal to $6$.  Hence we have:
$$3x^2+3=6$$
Solving for $x$, we see that $x=\pm1$. You found the first point $(1,8)$. Now just plug $-1$ into the original expression and you get $0$. Our second point is thus, $(-1,0)$. 
A: Firstly, solve $$(x^3+3x+4)'=6$$
I got $(1,8)$ and $(-1,0).$
