# 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.

• Welcome to MSE! How did you find the first point? Can you use the same method to find the second point? – Robert Howard Feb 23 at 18:47

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 $$(-1,0)$$.

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, $$(-1,0)$$.

• Thank you for the detailed description! So, to find the second coordinate one must use the first coordinates x and use it negatively in the second coordinates x to find out y? Is that correct? – Umehra Arfeen Feb 23 at 19:03
• You are welcome! If I understand what you are saying correctly, the fact that you use the first coordinate's negative x value is just because the derivative is a quadratic which has a positive and negative solution. This might not always be the case, I would learn the reason why we used -1. For the second part or your comment, just plug the "second x" value into the original function so you can find the appropriate y value. I hope this helps. – Gnumbertester Feb 23 at 19:22

Firstly, solve $$(x^3+3x+4)'=6$$ I got $$(1,8)$$ and $$(-1,0).$$

• Hey, sorry I still don't understand what I am supposed to do – Umehra Arfeen Feb 23 at 18:54
• @Umehra Arfeen You need to solve $x^2=1$, which gives $x=1$ or $x=-1$. – Michael Rozenberg Feb 23 at 18:55
• Take the derivative of $x^3+3x+4$ (also known as differentiating it) (for each term, multiply by the power, then reduce the power by $1$, i.e. $(ax^b)'=abx^{b-1}$. Then set the derivative equal to $6$, the gradient. – Rhys Hughes Feb 23 at 18:55