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I already was able to find out the first coordinate, which is $(1,8)$; however I can't seem to find the second one.

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  • $\begingroup$ Welcome to MSE! How did you find the first point? Can you use the same method to find the second point? $\endgroup$ – Robert Howard Feb 23 at 18:47
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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 $(-1,0)$.

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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, $(-1,0)$.

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  • $\begingroup$ 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? $\endgroup$ – Umehra Arfeen Feb 23 at 19:03
  • $\begingroup$ 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. $\endgroup$ – Gnumbertester Feb 23 at 19:22
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Firstly, solve $$(x^3+3x+4)'=6$$ I got $(1,8)$ and $(-1,0).$

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  • $\begingroup$ Hey, sorry I still don't understand what I am supposed to do $\endgroup$ – Umehra Arfeen Feb 23 at 18:54
  • $\begingroup$ @Umehra Arfeen You need to solve $x^2=1$, which gives $x=1$ or $x=-1$. $\endgroup$ – Michael Rozenberg Feb 23 at 18:55
  • $\begingroup$ 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. $\endgroup$ – Rhys Hughes Feb 23 at 18:55

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